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Hello Mistress Katie-Leigh

There is no cross multiplying here.  I do not like the term cross mutilpy at all anyway BUT

3/100*1200=N 

this is just     3/100*1200 = 36

and this is assigned the pronumeral N    (I mean it is called N)

N=36

Does that make sense. ?

Yes! Yes!

Thank u CPhil! And thank u too Lance even if it wasn't the answer I was looking for. :)

This is easy  !!!!

3 - 4 - 5     is one.......and all multiples of this relationship are triples...so

6 - 8 -10     [multiply the basic one by 2]

9 -12- 15    [ by 3]

12 - 16 - 20  [by 4]

15 - 20 - 25   [by 5]

18 - 24 - 30   [by 6]

Distance formula =  sqrt  [ (subtract x's)^2  + (subtract y's)^2 ]

a. (2, 5) and (7, 17)  =

sqrt[ (7 - 2)^2  + (17 - 5)^2]  =

sqrt [ 5^2    + 12^2]  =

sqrt [25 + 144] =

sqrt [169]  =   13 units

b. (-10, 1)and(-7, -3)  =

sqrt [ ( -7 - -10)^2 + ( -3 -1)^2]  =

sqrt [ (-7 + 10)^2 + (-4)^2]  =

sqrt[ 3^2  + (-4)^2]  =

sqrt [9 + 16]  =

sqrt [25]  =  5 units

Let AC = BC  = 1

And if ACB = 96  Then CAB = CBA  = 42

But ADB is a triangle and DAB  =  18  and  DBA = 30.......so ADB = 132

And using the Law of Cosines

AB^2  = AC^2 + BC^2  - 2 (AC)(BC)cos(96)   .....so.....

AB^2  = 1^2 + 1^2  - 2(1)(1)cos(96)

AB = sqrt [2 - 2cos(96)]

And by the Law of Sines

AD / sin(30)  = AB/sin(132)

AD / (1/2)  = sqrt[2 - 2cos(96)] sin(132)

AD  = sqrt(2 - 2cos(96)] / [2 sin(132)]     

AD  = [sqrt(2) (sqrt(1 - cos(96))] / [2sin(132)]

AD = [sqrt(2)/2] * [sqrt(1 - cos(96)]/ [sin(132)]

AD = [1/sqrt(2)] * [sqrt(1 - cos(96)] / [sin(132)]

AD = [ sqrt(1 - cos(96)/ sqrt(2)] / [sin(132)]

AD  = sqrt [(1 - cos(96)/ 2]  / sin(132)     

And by a trig identity  sqrt [(1 - cos(96)/ 2]  = sin(48)

So

AD  =  sin(48)/sin(132) ......   but sin(48)  = sin(132)    .....therefore....

AD = 1

And angle CAB  =   angle DAB + angle DAC 

42  = 18  +  angle DAC    ...therefore

Angle DAC  = 42 - 18   = 24

So....in triangle CAD

AD = CA = 1

And angle DAC = 24

So......since AD = CA....the angles opposite those sides are also  equal

Therefore, angle ADC = angle ACD = [180 - 24]/2  = 156/2  = 78

Therefore....Angle ACD  = 78

As follows:

.

25 = 5² a square,

27 = 3³ is a cube.

log(8)-log(9) = log(8/9)=log 0.8888889

Hi InjustaGod,

I have been super preoccupied and busy with other non-forum relaled things lately.  The forum is so big I have to allocate many hours every day to get through it all.  Anyway, I just have not had time.  I am going away for a week in a few days, although I do not know whether this will give me more or less time for the forum. :/

Anyway...

It might be a while -  sorry InjustaGod.  

Why? Do you like them? I mean is it that you like to see your name occasionally or do you look at some of the posts in there?  I do not get a great deal of feedback although I do know that some people look at them properly and I know some people glance at them but I am not sure how many people do or what the main attractions are.  :)

Anyway, thanks for asking.  :))  I appreciate that.    

Injustagod

if you are bored then read this.

I'm on, and bored. But our not on😤😤😤

Anything to do with Star Wars, except Kylo unmasked

Or...it's possible that they want you to identify the square and the cube from the given set of numbers.....if so.......

The square is 25, because 5 * 5  = 25

The cube is 27, because 3 * 3 * 3   =  27

Yes.

\(13^4 = 28561 \\ 14^4 = 38461 \\ ~\\ \dfrac{25561}{38461}\)

Hi Eloise :)

I think it is this that is wanted.

x     x^2     x^3

20   400      8000

21   441      9261

22   484    10648

23   529    12167

24   576    13824

25   625    15625

26   676    17576

27   729    19683

28   784    21952

29   841    24389

30   900    27000

That would depend what country you live in and what work entitlements you have.

Did you think everyone gets  the same ????

352 / 8  =   44  people

Most calculators will only give the decimal form.....WolframAlpha will display the exact value....here's the result : http://www.wolframalpha.com/input/?i=sin%28300%29

However.....if we remember some basic trig.......300°  is a 60°  reference angle in the 4th quadrant

And the sine of  60°   =  sqrt(3)/2   ......but.....the sine  is negative in this quadrant, so -sqrt(3)/2  is correct

Let d be the distance  to the water

The time to fall d distance is given by :

-d [meters] = (-4.9m/s^2)t^2

Rearranging this  and solving for t  we have

sqrt(d/4.9)  = t

And, the speed of sound  at 20C ≈ 343m/s

So......the time for the sound to travel from the water to the observer =

  [d meters] / 343m/s  =

d/343

So the total time is given by :

Time to fall + Time for sound return  = 6.58 sec

sqrt(d/4.9)  + d/343  =  6.58

Solving this for d, we have

d  ≈  179.712m

Thanks and congrats. You certainly got it right. Can you take my tests for me?!!!.

Just kidding.

Yours is more than cool Chris, It is brilliant.

It just needs one less 'S'   LOL

 Lim-->4 (x*sqrt(x)-8)/(x-4)  

\(\displaystyle\lim_{x\rightarrow4} \frac{ (x*\sqrt{x}-8)}{(x-4) }\\ =\displaystyle\lim_{x\rightarrow4} \frac{ (x*\sqrt{x}-8)}{(x-4) }\)

At this point I have learned from our guest.  Thankyou :)  

\(=\displaystyle\lim_{x\rightarrow4} \frac{ (x\sqrt{x}-8)}{(\sqrt{x}-2)( \sqrt{x}+2) }\\~\\~\\ \qquad \mbox{I need to get rid of the }\sqrt{x}-2\;\;\mbox{in the denominator}\\ \qquad \mbox{Now I used algebraic division to find that}\\ \qquad (x\sqrt{x}-8)\div (\sqrt{x}-2)=x+2\sqrt{x}+4\\\qquad so\\ \qquad(x\sqrt{x}-8)=(x+2\sqrt{x}+4) (\sqrt{x}-2)\\~\\~\\ =\displaystyle\lim_{x\rightarrow4} \frac{(x+2\sqrt{x}+4) (\sqrt{x}-2)}{(\sqrt{x}-2)( \sqrt{x}+2) }\\~\\ =\displaystyle\lim_{x\rightarrow4} \frac{(x+2\sqrt{x}+4) }{( \sqrt{x}+2) }\\~\\ =\frac{4+2*2+4}{2+2}\\ =3\)

Here it the graph - it has a hole in it at (4,3)

Use this formula: D=1/2.g.t^2=1/2.9.81.6.58^2=212.37 meters-dist. to the water.