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THANKYOU XD

huh :)

could you explain how to put one post card into two post boxes

box 1 - postcard

box2 - postcard

box3 - 2postcards

Oh I get you mr/ms Smart a**e :D

It is easy, you tear it in half and put a half into each letter box. Done :)))

I must be one of those talented people you spoke of :)

1 in 2 boxes and 2 in the other box.

There are amazingly talented persons in this world: could you explain how to put one post card into two post poxes.

Lets see

4 post cards

3 letter boxes.

All 4 in the same letter box, that is 3 ways.

3 in one and 1 in another and zero in the last

Well there is 4 ways to chose the odd postcard and then 3 boxes it can go in that is 4*3=12 then there is 2 boxes to chose from for the others so that is 2*12=12 ways

3 in one and 1 in another.  24 ways

2 in one and 2 in another and one empty.

Well there are 3 ways to choose the empty box. Then 4C2 ways to chose 2 that is 6 ways and they cand go in either of 2 boxes so that is 3*6*2 = 36 ways.

2 in one and 2 in another and one empty. = 36 ways.

1 in 2 boxes and 2 in the other box.

4C2=6 ways to chose the 2 together and they can go into one of the 3 boxes so that is 6*3=18ways

then each of the others can go in either remaining box, that is 2 ways. so altogether that is 18*2=36 ways

1 in 2 boxes and 2 in the other box​  =36 ways

So I get  3+24+36+36 = 99 ways

population density  =  ( number of people ) / ( number of square miles )

population density of AL   =  ( 4,779,736 people ) / ( 51,718 square miles )

                                          ≈  92.419  people per square mile

population density of MO  =  ( 5,998,927 people ) / ( 69,709 square miles )

                                           ≈  86.057  people per square mile

The average state population density is  87.4  people per square mile

So...Alabama has a more dense population than the national average,

and Missouri has a less dense population than the national average.

Hi Davis

\(log_e e=1\)

\(log_{10}10=1\\ \)

etc

A log is a power and the base can be whatever you want.

log base e is called a natural log. e is just a specific number just like pi is

e=2.718 ...

example 

\(log_{10}100=2\\ because\\ 10^2=100\)

l

Log(e)=1

OMG!! Paging Dr. Tud, emergency, come right away!

There is a new strain of CDD on here: CDD:BB –Contagious Dumbness Disease:Blarney B. This strain is a close variant of CDD:BS   Both of these strains mutated in a family of gnomes –E.F Gnomes in particular.    You may be patient zero. Try not to infect anyone else.  

Area of the big circle is pi*2^2 = 4pi

Area of the little cirlce is  pi*1^2 = pi

Now

\(\text{area of big = 4 area of little} - 4A_2+4A_1\\ 4\pi=4pi- 4A_2+4A_1\\ so\\ - 4A_2+4A_1=0\\ A_2=A_1\\ A_1-A_2=0 \)

I can explain better if you need me to :)

OR if anyone who is watching wants me too. :)

As long as the points aren't collinear, we can construct a circle that will pass through any three points......see the pic below.......

Call a, b the center of the circle

And let the circle pass through points (0,2), (2,2)  and touch point (1,0)

And we need to equate the distances from (a,b)  to (1,0)   and from (a,b) to (2,2)

(a-1)^2 + (b)^2  =  (a-2)^2 + (b - 2)^2

Expanding this we have

a^2 - 2a + 1 + b^2 = a^2 - 4a + 4 + b^2 - 4b + 4  simplify  

-2a + 1  =  -4a  + 4 - 4b + 4

2a + 4b  = 7     (1)

And then we can equate the distances from (a,b)  to (1,0)   and from (a,b) to (0,2)

(a-1)^2 + (b)^2 = (a)^2 + (b - 2)^2  simplify

a^2 - 2a + 1 + b^2  = a^2 + b^2 - 4b + 4

-2a + 4b = 3    (2) 

Adding (1)  and (2)  we have that

8b =10  →  b = 5/4

Subbing this back into (1)

2a + 4(5/4) = 7

2a + 5  = 7   →  a = 1

So the center of the required circle is ( 1, 5/4)

And the distance that this point is from (1, 0)  is just 5/4  -  (1.25) -   which is the radius

Here's a pic :

(a,b), represented by E, is the center........The circle passes through B, C  and is tangent to AD

This is a way to draw a circle that passes through two adjacent vertices of a square and is tangent to one side of the square....I think......

radius  =  diameter / 2

radius  =  side length of square / 2

radius  =  2 / 2

radius  =  1

Here is the actual hyperlink, but without the variables but, I considering it is a circle you can plug the pairs of variables and variables in the correct areas

http://www.davidmaroto.info/Four-Circles

if not... the image is calles Four+Circles.jpg but it has labels and two shaded reagions which are key...

if you know how I could possible insert shapes please let me know how to I can put to diagram on here the correct way... 

Answer and explination soon!!!

That's not much information to go on. Is your graphing thing broken? Whether or not, you can try using the graph software at geogebra.com and desmos.com. They are both free and work well!

Parallel lines have the same slope. We can find the slope of our line by finding the slope of the line

-8 + 52x + 97y  =  0           . Let's get this into slope intercept form. Subtract  97y  from both sides.

-8 + 52x  =  -97y                Divide through by  -97  .

\(\frac{8}{97}-\frac{52}{97}x=y \)                  Rearrange.

\(y=-\frac{52}{97}x+\frac8{97}\)               When the equation is in this form, we can see that the slope is  -\(\frac{52}{97}\)  .

So, the equation of the line with a slope of  -\(\frac{52}{97}\)  that passes through  (-78, -56)  is.....

y - -56  =   -\(\frac{52}{97}\)( x - -78 )

y + 56  =   -\(\frac{52}{97}\)( x + 78 )           This is in  " point-slope "  form. We can get it in slope-intercept form.

y + 56  =   -\(\frac{52}{97}\)x  -  \(\frac{4056}{97}\)          Subtract  56  from both sides of this equation.

y  =   -\(\frac{52}{97}\)x  -  \(\frac{9488}{97}\)

NOPE... actually i figuered out how to solve it over a good bit of time and thought...

you actually have to solve for y first using anither equation from the diagram...

6x+y=x+5y

subtract 5y from both sides

-4y=5x

divide by -4

y=5/4x plug that into the equation for y

(4x)+(6x+5/4x)=180˚ ->

45/4x=180˚ ->

720/45

x=16

plug that into the equation...

4(16)+6(16)+y=180˚

64+96+y=180˚

170+y=180˚

y=10

so...

(x y)=(16)(10)

(x y)=160

You have two variables, but only one equation.....therefore, you will have infinite answers for x, y

DELETED

Both statements are true

For instance, (0,1)  is on the line

This is both a solution to the equation  and this solution is a point on the line

The intersection of the medians divide  triangle ABC into 3 triangles with equal areas, triangles AGB, BGC and  AGC

And since the area of the large triangle is 6, the areas of each of these triangles is 2

So.....

Area of AGB  =  (`1/2)  (AB ) (GR)  = 2

 (`1/2)  (3 ) (GR)  = 2

(3) GR  = 4

GR = 4/3

Area of BGC  =  (`1/2)  (BC ) (GP)  = 2

 (`1/2)  (4 ) (GP)  = 2

(4) GP  = 4

GP = 1

Area of AGC  =  (1/2)  (AC ) (GQ)  = 2

 (`1/2)  (5 ) (GQ)  = 2

(5) GQ  = 4

GQ = 4/5

So.....GP + GQ + GR  =  1 + 4/5 +  4/3  =   47 / 15

At  2 times the distance   it's  1/(2)^3    = 1/8 of the strength   which is   8

At 3 times the distance it would be 1/(3^3)

and at 4 times the distance it would be 1/(4^3)     and so on.

 The radius of the cicumscribing circle is given by :

[ Product of the side lengrhs ]  /  [ 4 * Area of the triangle ]  =

[ 6 * 8 * 10 ] / [ 4 (1/2) * 6 * 8]  =    10 / 2  =  5 units

The radius of the incircle is given by 

[ 2*Area of the triangle] / [Perimeter]  = [ 2* (1/2) * 6 * 8] / [ 6 + 8 + 10]  = 48 / 24  = 2 units

So....the area of the circumcircle - area of the incircle  =

[pi] [5^2 - 2^2] =  pi * [25 - 4] = 21pi  units^2 ≈ 65.97 units^2

Since the strength of the magnet decreases as the inverse cube of the distance, therefore:

64 x 1/(20/10)^3 =

64 x 1/2^3 =

64 x 1/8 =8 gauss - the strength of the magnet at 20 cm.

This is a "Pythagorean Triple" right triangle....

Two of the altitudes are 7 and 24

The third altitude is  given  by [ 7 * 24] / 25  = 6.72

So the sum of the altitudes is  6.72 + 7 + 24 =  37.72

I agree, but I always put parentheses around mine, especially when you get to the complicated order of operations topics and pre-algebra and above that.