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By the way Chris you got it wrong.

The values of a and b are not known !!     ROF LOL

Hi Chris, TR and Heather,

I am totally lost on what you are saying TR and Heather.

What has some old post that has not even been address referenced got to do with the question here?

I guess Chris seems to have some idea that you are talking about but I don't.   

Maybe I should be on the forum more often then I might 'get' what is going on.   LOL 

Yes !
how do you calculate (Pn)?

Lorea and peps are friends they have orders from Sweden 200 boxes. To give the Americans all but each family is supposed to take19 boxes ,how many days are they going to give people the boxes ? '"...........from Guest

OK....

70%   .....there is an implied decimal after the "0".......we move the decimal two places to the left and drop the % sign........and "of" means "multiply"

So we have

.70 * 80   =

56

70% of 80

Human.....go ahead and post a problem so I can see what you're talking about...

Is there some question associated with this???

I'm having a tough time envisioning what you decscribe........can you post a pic ???

2.5(-3^2)-8(-3)-1.5     first, work inside the parentheses and realize that -3^2  = -9

2.5(-9)  -8(-3) - 1.5   next .... we perform any multiplications/divisions

-22.5 + 24 - 1.5  =

1.5   - 1.5  =

0 

Using PEMDAS, exponentiation takes priority over negation, so most technical calculators will treat -3^2 as -(3^2) = -9 not +9.  This explains why most calculators produce a result of 0 for the given expression.

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It means draw a shape (A, say) around another (B) such that A touches B but does not cut through it.  

So, in MATHBITCH's example the circle is circumscribed around the triangle.

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You could use the Desmos online graphing facility- see below:

Same signs....just add the integers and keep the sign

Example :    -21  +  - 28   =   -21  - 28   =   -[21 + 28]  = -49  ....note......we had the same signs, so we just added and kept the sign....

Different signs.......

Drop the signs......take the larger integer  and subtract the smaller.......append the original sign on the larger integer to your answer

Example :

-21  + 6

Drop  the signs  =    21, 6

Take the larger integer and subtract the smaller   21  - 6   = 15

Append the original sign on the larger integer to your answer......21 is the larger and it had a negative sign, originally.....so.......append this sign to the answer in the previous step

So....the answer  is :    -15

Another example:

-3 + 24

3, 24

24 - 3   = 21

The larger integer had a positive sign, so.....we just append that sign to the answer in the previous step   =  +21 .....  [which is the same as 21]

I don't see a picture, but I believe I know what you want.

7- horizontal in each  row         5 - vertical  in each column 

The number of horizontal toothpicks  =

(number of vertical toothpicks in each column + 1) * (number of horizontal toothpicks) =

( 6) * (7 ) = 42

The number of vertical toothpicks =

(number of horizontal toothpicks in each row + 1) * (number of vertical toothpicks) = 

(8) * (5) = 40

So.....we need  42 + 40   = 82 toothpicks

Number of dots (nodes)  =

(number of vertical toothpicks in each column + 1) * (number of horizontal toothpicks in each row + 1) = (6) * (8)  = 48 dots (nodes)

With 10 horizontal and 20 wertical

# of horizontal toothpicks = (21)* 10   = 210

# of vertical toothpicks  = (11) * 20 = 220

# of total toothpicks = 210 + 220  = 430

# of dots (nodes)  = (21) * (11)  = 231

We can extend this idea to any horizontal/vertical toothpick array.

Let h be the number of horizontal toothpicks and v the number of vertical

# of horizontal toothpicks  = (v + 1) * h

# of vertical toothpicks = (h + 1) * v

# of total toothpicks  = (v + 1) * h  + (h + 1) * v

# of dots (nodal points)  = (v + 1) * (h + 1)

cos2x - cosx = 0 i need steps....

rearrange

\(\begin{array}{rcll} \cos{(2x)} - \cos{(x)} &=& 0 \\ \cos{(2x)} &=& \cos{(x)} \\\\ \boxed{~ \cos{(\varphi)} = \cos{(-\varphi)}~}\\\\ \end{array}\)

We have 4 variations: {nl} \(\begin{array}{lrclrcl} (1) & \cos{(2x)} &=& \cos{(x)} \qquad \Rightarrow &\qquad 2x \pm 2k_1\pi &=& x \pm 2k_2\pi\\ (2) & \cos{(-2x)} &=& \cos{(x)} \qquad \Rightarrow &\qquad -2x \pm 2k_1\pi &=& x \pm 2k_2\pi\\ (3) & \cos{(2x)} &=& \cos{(-x)} \qquad \Rightarrow &\qquad 2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\\ (4) & \cos{(-2x)} &=& \cos{(-x)} \qquad \Rightarrow &\qquad -2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\\\\ (1) & 2x \pm 2k_1\pi &=& x \pm 2k_2\pi\qquad \Rightarrow &\qquad x &=& 0 \pm 2k\pi\\ (2) & -2x \pm 2k_1\pi &=& x \pm 2k_2\pi \qquad \Rightarrow &\qquad -3x &=& 0 \pm 2k\pi\\ (3) & 2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\qquad \Rightarrow &\qquad 3x &=& 0 \pm 2k\pi\\ (4) & -2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\qquad \Rightarrow &\qquad -x &=& 0 \pm 2k\pi\\\\ & \Rightarrow x &=& \pm 2k\pi \\ & \Rightarrow x &=& \pm \frac{2}{3}k\pi \\ \end{array}\)

We can put together  \(x = \pm \frac{2}{3}k\pi \qquad k \in N_0\)

I left out our physics people off my Credit list.  Alan is a Physicist and his posts sometimes blows my mind but Nauseated and Dragonlance as well as one or more guests often answer physics questions.  Physics is fascintating.  If there was an infinite amount of time I would definitely put effort into understanding physics.  :))

I have probably left some other great high end answerers off my list.... sorry.

Very nice, DarkBlaze.....!!!!

Yep, CPhill's answer id the same as mine :)

I expect you are right rarinstraw but that is the beauty of this place.

Almost all of us see mathematics here that is new or unknown to them.  I regularly see things that Alan, Heureka, CPhill, Bertie and Geno3141 put up on the forum that are either totally beyond me or that I can learn from.

I love that.

You are getting a glimpse of what is to come.  Little bits of it you can run with and learn from if you choose to.  Plus of course we are here if you do not understand your classwork :)

cos2x - cosx = 0       using an identity, we have

2cos^2x - 1 - cosx  = 0   rearrange

2cos^2x - cosx - 1  =   0     factor

(2cosx + 1) (cosx - 1)  = 0

Setting each factor to 0, we have

2cosx + 1  = 0     subtract 1 from each side

2cosx  = -1      divide by 2 on each side

cos x  = -1/2       and this happens at x = 120° ± n360°   and at x = 240°  ± n360°  where n is a positive integer

For the other factor, we have

cos x - 1   = 0   add 1 to both sides

cos x = 1       and this is true at x = 0° ± n360°    where n is a positive integer

Here's a graph : https://www.desmos.com/calculator/bza7s6b83u

st/6r= 32(45)/6(5)

(32*45)/(6*5) = 48

Let's just say this. If you are having problems with suicidal thoughts or actions, call this number: 1 (800) 273-8255

Thank you for apologizing as well

90% of all the math there is beyond me, at least, I think it is.

I'm sorry it was uncalled for. But she posts a lot of 'goodbye' messages.