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Gala2k [10]
3 years ago
12

Using the figure below, identify two line segments that complete the given expression: Given: FLAG is GXYZ is XY | | Segment GF

Segment AL Segment GZ Segment GX Segment FL
Mathematics
2 answers:
iogann1982 [59]3 years ago
4 0

The answer i think is Segment GZ.

ikadub [295]3 years ago
4 0

Answer:

i know there are 4 but i could only find three the 3 i found wher   gf gz fl there is one more tho

Step-by-step explanation:


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1. What is the fourth term in the expansion of (x + 4y)^6?
aev [14]
For the fourth term r = 3
Fourth term = 6C3(x)^(6 - 3) (4y)^3 = 20x^3 (64y^3) = 1,280x^3y^3
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This is the 6th time i've asked this question. i'm so desperate, please help me!
OverLord2011 [107]

Answer:

Look at explanation.

Step-by-step explanation:

For the first four, take the first term and add the common difference to it to get the second term. Then add the common difference to the second term to get the third term, and so on until the 5th term. If the common difference is negative, you subtract it instead of adding it.

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A B C OR D PLEASE HELP !!!!!!!!!!!!!!!!!!!!!!!!!!! I REALLY APPRECIATE
ddd [48]

Answer: B


Step-by-step explanation: Okay so you add 6m and 1m to get 7m for that entire side. Then you have 3m on each side that is slanted. You finally add 5m.

7m + 3m + 3m + 5m = 18m^2


4 0
3 years ago
How many different ways can you make 82 cents using current u.s. currency
andrew11 [14]
 <span>You can probably just work it out. 

You need non-negative integer solutions to p+5n+10d+25q = 82. 

If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80. 


So this is the same as n + 2d + 5q ≤ 16 

So now you simply have to "crank out" the cases. 

Case q=0 [ n + 2d ≤ 16 ] 

Case (q=0,d=0) → n = 0 through 16 [17 possibilities] 
Case (q=0,d=1) → n = 0 through 14 [15 possibilities] 
... 
Case (q=0,d=7) → n = 0 through 2 [3 possibilities] 
Case (q=0,d=8) → n = 0 [1 possibility] 

Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81 

Case q=1 [ n + 2d ≤ 11 ] 
Case (q=1,d=0) → n = 0 through 11 [12] 
Case (q=1,d=1) → n = 0 through 9 [10] 
... 
Case (q=1,d=5) → n = 0 through 1 [2] 

Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42 

Case q=2 [ n + 2 ≤ 6 ] 
Case (q=2,d=0) → n = 0 through 6 [7] 
Case (q=2,d=1) → n = 0 through 4 [5] 
Case (q=2,d=2) → n = 0 through 2 [3] 
Case (q=2,d=3) → n = 0 [1] 

Total from case q=2: 1 + 3 + 5 + 7 = 16 

Case q=3 [ n + 2d ≤ 1 ] 
Here d must be 0, so there is only the case: 
Case (q=3,d=0) → n = 0 through 1 [2] 

So the case q=3 only has 2. 

Grand total: 2 + 16 + 42 + 81 = 141 </span>
3 0
3 years ago
Simplify x^3 y ^4 / 3y^4
hjlf
We can cancel out the y^4 on the top and bottom of the fraction.

this gives us...

(x^3)/3
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