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we wish you well, Zegroes.

We have

x (x + 1)(x+2) (x + 3)  = 3024

Instead of solving this algebraically...let's make an "educated" guess....since the factors are "close together,".....let's take the 4th root of 3024....

This equals  .... about 7.415...then, we might guess that the integers would be 7, 8, 9, 10

But this is impossible.....since the product would end in 0.....also.....the result is 5040....too big!!!

Tnen, the remaining guess is that 6,7,8,9 might work  .....6 x 7 x 8 x 9 = 3024

Voila!!!....there's the answer....!!!

Let p be....... "parallel lines"     Let q be......"they are equidistant"

I believe the Law of Detachment   would go something like this :

1) If p, then  q

2) Assume p

3) Then, q

So let 1)  be......

"If you have parallel lines, the lines are equidistant"

Assume p

Then q follows...."They are equidistant"

The limit is 2:

.

(-3, 0), (6, 3), (9, -3)

Both coordinate values of every point are "expanded" by 3/2

So.......

(3/2) * (-2, 0)  = (-3, 0)

(3/2) * (4. 2)  =  (6, 3)

(3/2) * (6, -2) = (9, -3)

She has 4 cats......

(4/5) of them =  (4/5)4 = 16/5  = 3.2 cats

And 4/5 of 1 cat = .8 of a cat

So.....3.2 cats + .8 of another = 4

Algebraically.....let N be the number of cats......

Then.....

(4/5)N + 4/5  = N

4/5  = (1/5)N

And it's obvious that N = 4......

12/x + 12/x-1=-2   add 1 to both sides and simplify the left side

24/x  = -1       mutiply both sides by x and divide both sides by -1

-24 = x

I'll let you do thsi one......

To find the midpoint, (x, y}  = 

{ add the x coordinates of both points / 2 ,  add the y coordinates of both points / 2 }

18/12 = x/4     multiply both sides by 4

4(18/12)  =  4(3/2) = 12/2  = 6 = x

tan (62)  = x / 18

x = 18* tan (62)  = 33.85 m

Point Q lies on MN......

To see why,   join NQ.....then, by HL, triangles TQN and DQN are both congruent right triangles

thanks guys. I'll let you know the results tommorow

Yes I used to be like that but I spend so much time socializing with people in different time zones that I  usually finish for the day quite late.  

BUT

1 am is still pretty late.   

Yah, our hotel business will be booming!       

Good luck Zegroes,    

Here's hoping that things go well.....Z-Man   !!!!!

Well, T.R.  ......that certainly is an interesting method of analysis.....even with "barch's" and "gowipe's" assistance......LOL!!!

Let me make the process a little easier.......in any triangle, the greater angles are found opposite the greater sides......so......since the sides are (in order from largest to smallest)....24, 18, 12......  the  angles opposite those sides are (in order from largest to smallest)    P, Q, N

BTW.....I'll give T.R. 3 points just because CDD can happen to anyone......at any time....anywhere.......it's often undetectable by those who contract it.......

Don't worry, T.R.  ......most everyone on the forum has had a slight case of it.....innoculations are sometimes effective....but.....their potentcy and ability to fight off the dread disease usually wanes .......this equation is sometimes used to calculate the % of CDD remaining in one's system, x hours after a they have made a hideous mistake in answering a question......

%CDD =  log(log (ln e^pi)) + pi^e

The less "pi" involved -  the better........cake seems to have no effect on the formula.....so....."eat up" !!!!

The answer is: Error. It doesn't make any sence.

I just answered this one here......

34117178776

We already have the base area, DB........we just need to multiply this by the height....

So we have......

(3.5)in^2 x  (5) in =   17.5 in^3

I agree, Anonymous.......

Using the Law of Sines, we have

sin 123 / 43  = sin A / 19      .....therefore....

Sin A = (19sin(123)/ 43)  .......and using the sine inverse, we have.....

sin^-1 [19sin(123)/ 43]  = A =  about 21.75°

The polar form can be written as....  z = r (cos Θ + i*sinΘ )

r = √[(4)^2 + (-4)^2]  = √[16 + 16] = 4√2

And  Θ   =  tan^-1(-4/4) = tan^-1(-1)  = -pi/4    ...so we have

z = 4√2 ( cos (-pi/4) +  i*sin (-pi/4) )  

..... note that this is the principal value......any angle of the form -pi/4  +  2pi (n)  is also possible, where n is an integer