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m = 6 kg

F= 2.4 N

x = xo + vo t + 1/2 a t^2       x0 = 0   vo = velocity original = o   so:

x = 1/2 a t^2        x = 2 (given)       (1)

Now      a = (change in v) / (change in t) = 1.2 m/s / t  (2)   Substitute this into (1)

2 m = 1/2 a t^2

2m = 1/2 (1.2 m/s / t) (t^2)

2 = .6 t        so t = 3.333 sec

Remember   F= ma    so: 

                    F = .6 * a           and    a = 1.2/t    and   t  = 3.333

                    F=   .6 *  1.2/3.333 = 2.16 N = Net Force (East)      

Frictional force is     2.4 - 2.16 N =  .24 N  West       (or  -.24 N East)

Thank you very much

The VERTICAL component of the 180 km/hr @ 32 degrees will be:

180 sin (32) = 95.385 km/hr

Now:

1000m = 1 km

1 km / 95.385 km/hr  =  .0105 hr = .629 min = 37.74 sec

Thanks x^2,

You do not need to take down your original post, you didn't say anything bad :)

Hi, Melody

Thank you for the compliment. You deserve one, too, especially for being constantly online since 2013! The other answerers do not receive enough thanks for their generosity, so I am glad you acknowledge their abilities. If you were not passionate about this forum, then it probably would not survive. The students run the forum; we provide the amazing feedback with this amazing community of people. Without the students, the forum would die. Cphill has a similar mentality to me. If I help one, then I think I have achieved my goal. 

The point system is a psychological yet social touch. It encourages me to return to this forum and watch it increase ever so slightly. It encourages me to return to this forum when I see a +5 next to my answer with a note from the questioner saying "Thank you!" 

You have achieved your goal for me; I will think about the "why" question now. 

I like the idea of "half answers" or some variation that enables the receiving end to think, learn, and interact more. Unfortunately, I do not know how to accomplish this. I can attempt to share ideas, but they are not guaranteed to be effective. 

I will continue to be an asset to this forum. I am glad you are, too. I am sorry if I sounded offended. I was not. I am actually considering taking down the previous post I made. 

Gavin I think I follow what you are saying and it sounds logical but a cone IS a pyramid.

It is a circular pyramid. So when you say a pyramid I think you are refering to a square or a triangular pyramid..?

\(\displaystyle \lim_{x\rightarrow ∞} \sqrt{\frac{16x}{4x+9}}\\ =\displaystyle \lim_{x\rightarrow ∞} \sqrt{\frac{16x\div x}{(4x+9)\div x}}\\ =\displaystyle \lim_{x\rightarrow ∞} \sqrt{\frac{16}{4+\frac{9}{x}}}\\ = \sqrt{\frac{16}{4+0}}\\ =\sqrt4\\ =2 \)

Hi X-squared,

My observations was not aimed at anyone in particular and certainly not at you. You have been providing great answers for a long time and I am very grateful to you.  I do not contribute as much as I used to but I continue to be passionate about this forum and I know that some question askers learn a great deal from the fabulous answers that you and others give.  It is these students that inspire us and make this whole site worthwhile. 

CPhill has told me that if he teaches even just one student and helps to foster in them a great love of mathematics that that is all the reward he needs.  (I hope you do not mind me paraphrasing Chris)

X-squared this is nothing wrong with you being pleased that your own score is increasing. I like seeing mine increase too. It is all a part of the 'game'. Do not criticize yourself for it. :)  Also I do not intend to be critical of people who pick questions that they personally are interested in. There is nothing wrong with that either. 

I just wanted answerers to think a little more before they answer. That is all.

For my own part I do not intend to answer questions if they are posted in multiples. But sometimes the question will interest me and I will answer anyway.  I will also make more of an attempt to only half answer questions but like many of you, it is always very tempting to give whole answers, we are precise people, we like to finish what we start. 

Plus sometimes I get frustrated because I give a half answer for a very good reason, and then someone else comes in and completes it. I do not mind if it is a child who is learning but it does frustrate me when it is someone with higher abilities.

I'd really like it if we could encourage the question askers to interact with us more  It is great when question askers challenge  or question my answers. It shows me that someone at the other end is really trying to learn!

I expect all you wonderful answerers feel the same way.

I will finish now with a huge thank you to all the wonderful people who put so much effort into teaching people here. And also a big thanks to those students who seriously learn from us and make our time here all worth while.

I normally will not answer questions when there are several asked on the same post but I am making an exception here because a much respected member has asked me too.  (He wants to see how it can be answered for himself)

From a point interior to a regular hexagon, perpendiculars drawn to the sides on the hexagon have lengths, in order of size of 3, 6, 7.5, 13.5, 15 and 18. How long is the side of this hexagon?

A hexagon has got 6 sides. Opposite sides are parallel. So the perpendicular from a point to one side will also be the perpendicular from the point to the opposite sides.

If you look at the given lengths you can see that 

3+18=21

6+15=21

7.5+13.5=21

So 21 is the perpendicular distance between the parallel sides.

So the centre of the hexagon is 10.5 perpendicular units from each side.

The angle sum of a hexagon is 180*(6-2) = 720 degrees

Each internal angle is 720 / 6 = 120 degrees.

Half of each internal angle is 60 degrees.

So this is what we have.

\(tan60^\circ=\frac{10.5}{x}\\ \sqrt3=\frac{10.5}{x}\\ x=\frac{10.5}{\sqrt3}\\ x=\frac{21}{2\sqrt3}\\ x=\frac{21\sqrt3}{2*3}\\ x=\frac{7\sqrt3}{2}\\ \text {the side length is 2 * x}\\ Side\;length\;=7\sqrt3\;units \\ Side\;length\;\approx 12.124\;units \\ \)

Here is the pic with the original figures. This pic was not needed to answer this question but it may help you to understand better.

Express \(x^2-x-2 \text{ as }(x+1)(x-2)\)

Multiply every term by \((x+1)(x-2) \text{ to get } 3x(x-2)+5(x+1)=15\)

Collect terms:  \(3x^2-x-10=0\)

Solve the quadratic to get \(x=-\frac{5}{3}\text{ and }x=2\)

2 is an extraneous solution (try substituting 2 into the original equation and see what you get!).

This post challenges my overall mentality when I answer someone's question, but it definitely shifts it for the better.

"Are you seriously trying to help the asker to learn?"

\(\frac12\le\frac x8\le\frac79\)

We commonize the denominator:

\(\frac{36}{72}\le\frac{9x}{72}\le\frac{56}{72}\)

This works out when x=4, 5, and 6

I hope this helped,

Gavin

Please don't just type in the answer, read the solution and try to understand the problem. Try to do other problems like this on your own. Then you can actually improve. 

There are two methods to do this problem.

Method 1: Pattern Finding

There are 9 palindromes with 2 digits 

There are 90 palindromes with 3 digits

There are 90 palindromes with 4 digits

Method 2: Counting

The 4 digit palindrome is in this format:

ABBA

There are 9 choices for A

10 for B

9*10=90

I hope this helped, 

Gavin

I totally agree with you.

One way to remedy this issue is to post the full solution but leave out the last steps, so the person asking the question can have time to read the solution and must understand it to figure out the answer. 

Gavin.

I agree with everything here. 

If the individual making the question needs a direct answer, I'll give them a direct answer ( Which isn't generally helping them, and it is what they want) but if they really need help I'll to do it step by step and do my best to help them. 

(5C1+5C2+5C3) * (6C2+6C3+6C4)

ASHRA

You are fortunate that CPhill did your homework for you. 

I thought you were rude and lazy when you posted three questions all at once that were obviously for homework.

NOW you have confirmed that you are rude by failing to acknowledge that it was CPhill who answered your questions (on a new post)

YOU "Found" the answers here because CPhill had already posted them!

You didn't even bother to thank him!! 

A.

x^2  + y^2 - 4x -6y -12  = 0 

x^2  - 4x   + y^2   - 6y    = 12              complete the square on x and y

x^2  - 4x  + 4  + y^2  - 6y  + 9  =  12 + 4  + 9        factor the left side

(x -  2)^2  +  ( y - 3)^2   =  25

B.  This is a circle  centered at   ( 2, 3)   with a radius of 5

C. Here's the graph :  https://www.desmos.com/calculator/e4mzndmixp

4. Expand

\(ay^3+by^2+cy+d=3x^3-8x^2+7\\ \ \\ a(x-2)^3+b(x-2)^2+c(x-2)+d=3x^3-8x^2+7\\ \ \\ ax^3−6ax^2+12ax−8a+bx^2−4bx+4b+cx-2c+d=3x^3-8x^2+7\\ \ \\ ax^3-6ax^2+bx^2+12ax-4bx+cx-8a+4b-2c+d=3x^3-8x^2+7 \)

We get:

\(ax^3=3x^3 \\ bx^2-6ax^2=-8x^2\\ 12ax-4bx+cx=0\\ 4b-8a-2c+d=7\)

We get the systems:

\(a=3\\ b-6a=-8\\ 12a-4b+c=0\\ 4b-8a-2c+d=7\)

\(a=3\\ b-6\cdot3=-8\Rightarrow b=10\\ 12\cdot3-4\cdot10+c=0\Rightarrow 36-40+c=0 \Rightarrow c=4\\ 4\cdot10-8\cdot3-2\cdot4+d=7\Rightarrow40-24-8+d=7 \Rightarrow d=-1\)

\(a+b+c+d=3+10+4-1=\boxed{16}\)

I hope this helped,

Gavin