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The probability is 15/28.

PLease HELPPPPPPPPPP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

its cus the question is soooooo long

if you click that website its the question there

What is your "specific question" that you have problems with? Can't you just type it?!!

16 x 15 x 12 = 2,880

2,880 / 12   = 240

240 = (1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240) >>Total = 20

240 is the smallest positive integer.

In other words,

the cost of 1 inch is = 8.75/36 

gives some value, multiply it by 22 inches and that is the answer :)

This is a good question, it has made me think.  

Our math club has 20 members, among which we have 3 officers: President, Vice President, and Treasurer.

However, one member, Ali, hates another member, Brenda.

How many ways can we fill the offices if Ali refuses to serve as an officer if Brenda is also an officer?

(No person is allowed to hold more than one office.)

\((20\times19\times18)-(3\times2\times18) = 6840 - 108 = 6732\)

What is the greatest possible number of points of intersection for eight distinct lines in a plane?

Let \(\mathbf{n}\) is the number of lines.

The greatest possible number of points of intersection for \(\mathbf{n}\) distinct lines in a plane is : \(\dbinom{n}{2}\)

Here \(n = 8\):

\(\begin{array}{|rcll|} \hline && \dbinom{n}{2} \\\\ &=& \dbinom{8}{2} \\\\ &=& \dfrac{8}{2}\cdot \dfrac{7}{1} \\\\ &=& \mathbf{28} \\ \hline \end{array} \)

let the legs be x units long

Use pythagoras's theorum to solve for x.

Then you can find the area. 

Oh, also mention here if you answer it because otherwise we will probably not see you answer.

I have just now made an attempt on the original thread.

Please continue it there.

Asked again here

I am not sure but this is what i am thinking

Any on of the 7 can go in the middle 

Place one more is a specific spot - Say the red one 

There are 5 more places to fill so that is 5!

Now if the middle one and one side one is fixed then there is only one axis of symmetry that counts,  I mean the one that goes through the centre one and the red one.

So maybe the answer is

\(\frac{7*5!}{2}=7*3*4*5=420\)

I am ot sure though.

Eighty years ago, someone planted a tree in Brooklyn. It’s still there.

Someone wrote a book about it:  A Tree Grows in Brooklyn

): im most likely gonna "destroy" up on my test tomorrow then lol 

but no lol i made the question up :') 


thanks tho for your help again 

If it goes in ABDC, then your proof should be right.

The diagonal part, I am not so sure, you might have to get someone who is experienced in Honors Geometry.

Do you have the exact problem? if you can, try to post a picture.

I might have to go soon, sorry, I hope CPhill or Melody or someone educated can help!

yes it goes in the order ABCD except swap d and c 

Of course its a parrelelogram, but is it in the order ABCD?
 

Because your correspondences dont seem to make sense.

It is supposed to go in alphabetical order like the image above.

Look at your proof, then look at the image above, does it make sense?

yea it's a parallelogram 

also they aren't equal in length though,,, i thought for a parallelogram i had to prove diagonals non congruent (because it might interfere with the def. of rectangle) and opposite sides parallel (which is the same for rectangles except rectangles have congruent diagonals) 

maybe i have to prove they bisect each other as well as to the other proofs though,,, 

so would i use the two diagonals to find where they intersect and the midpoint is the proof? 

Is this theorem given?

" _{opposite pair of sides are parallel lines and diagonals are not perpendicular "}

I am kind of confused, for your proof, is it based on parellelogram ABCD?

If it is based on ABCD, you have to check the corresponding segments.

Also, \(|\overline{CB}|\neq|\overline{DA}|\) is incorrect I think. Because that is saying that they are not equal in LENGTH. You want to prove that they are NOT perpendicular.

So you have to do this I think. \(|\overline{CB}|\not\perp|\overline{DA}|\)

what just happened....

yes, you did but did i do it right or wrong.....