All Answers 358090 Answers

Y = 0.000085685866425

\(F_s = \mu_sN\), where \(F_s = 4.11\) and \(\mu_s = 0.325\)

\(4.11 = (0.325)N\)

\(N = \frac{4.11}{0.325}\)

\(N = 12.65\)

The number you add is the complex number a + bi, which here is -1 + 12i.

Oops sorry I got it

Force required = µs*weight = 0125*0.382*9.8 N

So what number do you have to add?

Let the unknown (complex) number be a + bi

Then  3 - 5i + a + bi = 2 + 7i

Subtract 3 - 5i from both sides:   a + bi = 2 + 7i - 3 + 5i or a + bi = -1 + 12i  

Equating real and imaginery parts we have:  a = -1,  b = 12. 

. 

300 ( 1 + .03)^4  = $337.65

how much force does it take to push a 62.0 kg dresser at a constant speed across a carpet with µk=0.238. (unit=N)

Your formula is not accurate.

FV=0;N=5*365; R=0.048/365;PV=2500; FV=PV*(1 + R)^N

FV =$ 3,178.07 - The balance in the acct. after 5 years.

3.05 x 9.81 N =3.05*9.81 = 29.9205 N   (Normal force)

29.9205 N x .250 = 29.9205*.25 = 7.480125 N Frictional force

a 3.05 kg block is sliding across the ground with μk=0.250. what is the friction force?

Hmmmmmm.....    I think I agree with Anthrax on this one.  The question did not ask for the NORMAL force on the hand....it asks for the magnitude of the  NET force on the hand.....  Ben, you even stated that the 130 N results in increased frictional FORCES on the hand.....   the RESULTANT force acting on the hand is not just the normal force of 1127 , but ALSO the horizontal force of 130 N and these two forces DO add as Anthrax calculated.....and they act in the direction of the hypotenuse (i.e. angled....not NORMAL).....   My one up vote goes with Anthrax.

  My 2 cents worth

Thank you very much.

Just tidying up, (I'm not guest #2).

The second value of a is

\(\displaystyle \frac{1}{9u}+u-\frac{1}{3}, \text{ where }\quad u=\frac{(3\sqrt{57}+23)^{1/3}}{3\times4^{1/3}}\).

So, not rational.

Prove:

\(1-8sin^2\theta cos^2\theta=cos(4\theta)\\ LHS=1-2(2sin\theta cos\theta)^2\\ LHS=1-2sin^2(2\theta)\\ LHS=1-sin^2(2\theta)-sin^2(2\theta)\\ LHS=cos^2(2\theta)-sin^2(2\theta)\\ LHS=cos(4\theta)\\ LHS=RHS \qquad\\ QED \)

Yes I accept that   

Sorry.

I wasn't trying to be mean or disrespect him. My reply does sound a bit harsh in retrospect haha. I didn't mean anything bad by it. I hope he knows that. Just trying to help and give my insight.

How many 3-digit numbers between 100 and 999 are there with the property that each digit, from the left, beginning with the 2nd is > than the digit before it, and the 3rd is > than the 2nd, such as : 123, 234, 456....etc? And how many more 6-digit such numbers are there between 100,000 and 999,999? Any help would be great. Thank you. 

It stands to reason that 0 cannot be one of the digits and all the digits must be different. 

How many ways can 3 digits be chosen from 9, that would be 9C3=84

For each of those triples there is only one way that they can be placed in ascending order

So the answer is 84

You can now work out the number of 6 digit combinations for yourself.  It will be done in the same way.

Thanks for your answers Anthrax and Ben.  

But...

Ben, it would have sounded an aweful lot nicer if you had started with something like:

'Antrax, I do not think your answer is right because .......'

If you had actually asked the question, which you didn't, then I'd also suggest that you started with 'Thanks'

Opening with

'You are wrong', 

or   'Your result is incorrect' 

or similar statements is a great way to get our esteemed answerers off side, and there is nothing acheived by doing that.  

This result is incorrect. The vertical and horizontal forces are separate and do not add in that manner. Imagine standing next to a wall and start pushing perfectly horizontally on that wall, do you feel heavier? If you push hard enough on a wall will your legs eventually give out from too much force? I hope not. The vertical normal force on the hand is only the 1127 N. The 130N that is applied horizontally would result in a torque force on the hand, and an increase in frictional force on the hand, not actual pushing forces. What you were calculating is the magnitude of the total force on the acrobat, not the force on the hand.

F = ma

F = 115 x 9.8m/s²

F = 1127N

This is due to Newton's first law stating that force equals mass times acceleration, the acceleration being the gravity of earth, 9.8m/s/s, and the mass being the acrobats ‘weight’ although it is actually his mass. The 130N applied to the horizontal wall is of no consequence to the force that is applied to the ground, while it would apply a torque force to his hand, this would be in Nm^-1 and incalculable given the current information.

So, our current evidence that supports a difference exists between the first group and the second group is the difference in means being 3. So, looking at a line plot of random simulations, we want 3 to be an unlikely result, thereby showing that the difference in means did not happen by random chance. However, we do not see that in this case. 7/10 of the random simulations resulted in a difference between means in the two groups of or greater than three, so therefore the difference in the means of the two groups is not significant.

Incorrecto; this sentence contains no command form of the verb 'comer', instead it just reads 'you eat a lot' and I feel personally attacked tbh.

a) I will be substituting for x and simplifying.

\(\frac{4x^3+2x^2+6x+7}{2x+1} = 2x^2 + 3 + \frac{4}{2x+1}\)

\(\frac{4(1000)+2(100)+6(10)+7}{2(10)+1} = 2(100) + 3 + \frac{4}{2(10)+1}\)

\(\frac{4267}{21} = \frac{4267}{21}\)

(b) When plugging in \(x=-\frac{1}{2}\) to the denominators \(2x+1\) on both sides, we will end up dividing by zero, an undefined value.

(c) There's really two ways to do this. One is multiplying both sides by \(2x+1\) and showing the results. The other is dividing \(4x^3+2x^2+6x+7\) by \(2x+1\) through long division to show that it equals \(2x^2+3+\frac{3}{2x+1}\). However, I happen to hate long division with polynomials. So Ima do it tha otha way.

\(\frac{4x^3+2x^2+6x+7}{2x+1} = 2x^2 + 3 + \frac{4}{2x+1}\)

\(4x^3+2x^2+6x+7 = (2x+1) (2x^2 + 3 + \frac{4}{2x+1})\)

\(4x^3+2x^2+6x+7 = 4x^3+2x^2+6x+7\)

The right side multiplication looks intimidating, but it actually multiplies out that easily in one step.

Oh, and, see (b) for why it doesn't work for \(x = -\frac{1}{2}\)