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I think it is possible...

Let \(\angle AXE = \angle CYX = x\)

We have the equation: \(x = 3x-98\)

Solving for x, we find that \(x = 49\).

Now we have: \( \angle AXE = \angle BXY = \color{brown}\boxed{49}\)

What am I missing?

And the question is?

a  =  6

a² + b²  =  c²     --->     6² + b²  =  c²     --->     36 + b²  =  c²   

a + b + c  =  20     --->     6 + b + c  =  20     --->     b + c  =  14     --->    b  =  14 - c

36 + (14 - c)²  =  c²     --->     36 + 196 - 28c + c²  =  c²     --->     232 - 28c  =  0

                                    --->     28c  =  232     --->     c  =  58/7 

Part A: Create two variables and determine which is dependent and which is independent for this situation. (4 points)

Part B: List three ordered pairs to show the values of the two variables. (3 points)

Part C: Write an equation to represent the relationship between the two variables

Part A: Create two variables and determine which is dependent and which is independent for this situation. (4 points)

Part B: List three ordered pairs to show the values of the two variables. (3 points)

Part C: Write an equation to represent the relationship between the two variables

If both have the same  number at the end  then  each  has   (272/2)   = 136

Let  E  be the number that Eugene started with and F what Frank started with

And   F  = 272 - E

Eugene gives away  (1/3) of his

And from Frank he gets   (1/5) ( F  + (1/3) E )  =   (1/5) ( 272  - E + (1/3)E )     (2)     

So  at the end,   what  Eugene started with  - what he gave away + what he got from Frank  = 136

So

E  - (1/3)E  +  (1/5) [ (272 - E + (1/3)E  ]  = 136   

(2/3)E +  272/5 + (1/5)(-2/3)E  = 136

(2/3)E - (2/15)E  =   136   - 272/5

[ 8/15] E  =  408 / 5

8E  =  1224

E = 153 = stamps that  Eugene started with

And Frank started with 119

Proof

Eugene gives  (1/3) to Frank = (153) / 3  = 51

Eugene now has   102

And now Frank has  119 + 51   = 170

He gives 1/5  of these to Eugene  = 34

Eugene now has    102  + 34  = 136

Frank now has  119 + 51  - 34  =  136   

thx you guys

5x + 3y  =  101

x        y             xy

19      2             38

16      7           112

13      12         156

10      17         170

  7      22         154   

\(2+2=4\)

I might be wrong here, I probably need one of the mods to check my work...

-Vinculum

Also,

"Given that all the numbers from 1 to N exist, the number N+1 is defined as -

N+1=N∪{N}N+1=N∪{N}

In other words we can say that the operator “+1” does nothing but unions the largest set smaller than the set we are operating upon with the set containing this largest set smaller than the set we were operating upon. (This is just the definition of the equation above :P )

We define 0 as an empty set,

0={}=Φ0={}=Φ

Now we recursively define the next numbers -

1=0∪{0}=Φ∪{0}={0}1=0∪{0}=Φ∪{0}={0}

2=1∪{1}={0}∪{1}={0,1}2=1∪{1}={0}∪{1}={0,1}

3=2∪{2}={0,1}∪{1}={0,1,2}3=2∪{2}={0,1}∪{1}={0,1,2}

4=3∪{3}={0,1,2}∪{3}={0,1,2,3}4=3∪{3}={0,1,2}∪{3}={0,1,2,3}

and so on…….

We can actually this way prove that - N={0,1,2,3,...,N−1}N={0,1,2,3,...,N−1}

Now N+2=N+(1+1)=(N+1)+1=(N+1)∪{(N+1)}={0,1,2,3...,N−1}∪{N}∪{(N+1)}={0,1,2,3...,N+1}N+2=N+(1+1)=(N+1)+1=(N+1)∪{(N+1)}={0,1,2,3...,N−1}∪{N}∪{(N+1)}={0,1,2,3...,N+1}

A good question would be why did the operator “+2” become equal to “+(1+1)”, this is simply because we defined 2 in that way. Also why did the brackets associate? Simply because we are dealing with sets(that too without any non-trivial operators), we can do that. ;)

Now we just plug in N=2 and Voila!!

2+2={0,1,2,3}=42+2={0,1,2,3}=4

PS. Sorry for being so long, but this is how we do it."

Easier  to  first caculate the probability  that  no faces match  = 

[(6) (5) (4) (3)  ] / 6^4  =  360 / 1296 =  5/18

So the probability  that at least two faces match  =   1 - 5/18  =  13 / 18

You are missing angle measures. There is no way to solve it without more. Following my knowledge of proofs and just proving angles in a nutshell there is no way to do this....

If you can give us more angle measures I'll be more than happy to help! 

ILP

This is impossible

If   AB  and CD are parallel, then angles   AXE and CYX are equal  since they are corresponding  angles

Assuming that she cooked one and left it, and got a different one and ate it whole, 

The answer would be 3...

If it she ate the one she cooked then it would be 4

-Vinculum

She has 4 eggs as one is just boiled. 5 total eggs minus the one that she ate

a^2 - b^2  =  32     

b^2 - c^2  >  0

If a^2  - b^2  =32, then a >b

And b > c

Let a =  2n + 1    and   b  =2n - 1     and   c =  2n - 3

So

(2n + 1) ^2  -  (2n  - 1)^2  = 32       simplify

4n^2  + 4n + 1   - 4n^2 + 4n  - 1   = 32

8n = 32

n = 4

c = 2(4)  - 3   = 5

b = 2(4) - 1   =  7

a = 2(4) + 1  =  9

So

b^2   - c^2   =  

7^2  -  5^2   =

49   -  25   = 

24

Are you sure you got the question right?

You want 4-digit integers and your set also has 4 digits, therefore all 4-digit permutations will have one 8 in them as follows:

{{2, 4, 6, 8}, {2, 4, 8, 6}, {2, 6, 4, 8}, {2, 6, 8, 4}, {2, 8, 4, 6}, {2, 8, 6, 4}, {4, 2, 6, 8}, {4, 2, 8, 6}, {4, 6, 2, 8}, {4, 6, 8, 2}, {4, 8, 2, 6}, {4, 8, 6, 2}, {6, 2, 4, 8}, {6, 2, 8, 4}, {6, 4, 2, 8}, {6, 4, 8, 2}, {6, 8, 2, 4}, {6, 8, 4, 2}, {8, 2, 4, 6}, {8, 2, 6, 4}, {8, 4, 2, 6}, {8, 4, 6, 2}, {8, 6, 2, 4}, {8, 6, 4, 2}}==24 permutations.

Note: Check your question carefully. I think there is something missing.