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It happens everywhere, I have to wait a few hours before I am reply to answers when it happens.  There is no one place it happens

To start this, let's create a ratio for the two triangles (the triangle you are trying to find, and the super large triangle). Taking the two  longer sides, we get that the scale is 13/65=1/5. Multiply 60 by 1/5, and we get that '?' is 12.

:)

Let $f(x) = 2x+1$ and $g(x) = f(2x) - 3$. What is $g(g(2))$?

To start this, we see that g(x)=f(2x)-3. So..., we can plug in 4 into f(x). This will give us 9. From there, we subtract 3 and get 6. Repeat the process, but plug in 12 for x in f(x). This will give is 23-3=20. Therefore, g(g(2))=\(\boxed{20}\)

Wheres CPhill on Leaderboard? :(

Press the little arrow on the bottom. Then CPhill will show up on the top. (prob because his score is too high, or he has not been on lots, lol)

:)

is that Aops? Cause if it is I can get u the answer

Here is Pascal's Triangle:

1   = R 0
1   1   = R 1
1   2   1   = R 2
1   3   3   1   = R 3
1   4   6   4   1   = R 4
1   5   10   10   5   1   = R 5
1   6   15   20   15   6   1   = R 6
1   7   21   35   35   21   7   1   = R 7
1   8   28   56   70   56   28   8   1   = R 8
1   9   36   84   126   126   84   36   9   1   = R 9
1   10   45   120   210   252   210   120   45   10   1   = R 10

The 10th row should have a sum of: 2^10 =1,024

That is correct!

66  166  266  366  466  566  606  616  626  636  646  656  660  661  662  663  664  665  666  667  668  669  676  686  696  766 >> Total =  26

Let's call one pack of gum X. 

The ratio of # of cherry to # of grape is .\(\\(\dfrac{20}{30} = \dfrac{2}{3})\) Using this, we have the following equality.  

\((\dfrac{20-X}{30} = \dfrac{20}{30 + 5X})\). The ratio to the left is the ratio of gum if chewman loses 1 pack of cherry, and the ratio to the right is the ratio of gum if chewman finds 5 packs of grape. 

Cross multiplying, we have \((20-X)(30+5X) = (20)(30) \Rightarrow (20-X)(30+5X) = 600\). Factoring, we have \(-5x^2 + 70x + 600 = 600 \Rightarrow -5x^2 + 70x = 0 \Rightarrow 5x^2 = 70x \Rightarrow \boxed{x = 14}\).

There are 14 pieces of gum in one pack of gum.

Yes sorry, the answer is 270.

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Did you post the same question yesterday? Look under here and tell us what is answer?

https://web2.0calc.com/questions/asap_12

The shortest distance is sqrt(7)/2.

Important: " ... where the order of my choices does matter "

If I understand the question correctly, you have only 15 choices:

1      1+2, 1+3, 1+4, 1+5, 1+6

2      2+3, 2+4, 2+5, 2+6

3      3+4, 3+5, 3+6

4      4+5, 4+6

5      5+6  

So, the answer is 15  

@melody if you can help me possibly ?

Multiply them together to get the product.....      1  ,   -8    ,   16

What would be the product of the constants and coefficients

Sorry, I didn't give you the link. Here it is:  https://www.geeksforgeeks.org/smallest-triangular-number-larger-than-p/

(x^2 -4x + 4 )  *  (x^2 + 4x +4 )

x^4 + 4x^3 + 4x^2       - 4x^3 - 16x^2 - 16x      + 4x^2 + 16x + 16

x^4 - 8x^2 + 16               coefficients  of   1  - 8    and constant of 16

Sorry for my typo     1/2 (x^2 - 14x) = 16

2x^2 - 20x +50 = 0

x^2 -10x + 25 = 0

(x-5)(x-5) = 0       x = 5

Solve for x:
1/2 (x^2 - 14 x) = 16

Multiply both sides by 2:
x^2 - 14 x = 32

Add 49 to both sides:
x^2 - 14 x + 49 = 81

Write the left hand side as a square:
(x - 7)^2 = 81

Take the square root of both sides:
x - 7 = 9 or x - 7 = -9

Add 7 to both sides:
x = 16 or x - 7 = -9

Add 7 to both sides:

 x = 16                    or          x = -2 [Throw this one out]

GO is the full length      16 is 1/2  of the full length

sooooo    1/2 (x^2 - 14) = 16

Go online to this page and learn how to find the nth triangular number. Click the "Python 3" code and see how it is done. Good luck.

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For the community: Please don't submit solutions to this problem. This is a homework problem for an online course that does not allow students to search for answers to specific problems outside of the school.

For the original poster: We realize that homework can be challenging. If you wish to receive help from the staff or other students, we encourage you to use the resources that the online classes provide. Please don't ask or search online for homework help. We understand that it's common in today's information age to look for resources online, and in some contexts, that's a great thing! However, it's against our Honor Code. You can ask for help on the message boards, and you can learn from the official solution after you submit your answer.