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a + b  = 7       

ab  = -18   

Find a^3 + b^2

Square both sides of (1)  a^2  + 2ab +  b^2  = 49   ⇒   a^2 + b^2  =  49 - 2ab  

a^3  + b^3  =    (a + b) (a^2 - ab + b^2)  

Subbing....we have that

a^3 + b^3 =  ( 7)  ( a^2 + b^2  - ab)

a^3 + b^3   =  ( 7) (49 - 2ab - ab)  

a^3 + b^3  =  (7) (49  - 3ab)

a^3 + b^3 =  (7) (49 - 3(-18) )

a^3 + b^3  =  7 ( 49 + 54)

a^3 + b^3 = 7 ( 103)

a^3  + b^3  =   721

Easily solved would be with a calculator

847+928 = 1775

o/w by hand

¹ 84¹7

   92 8

1775

M and N have to be as far away as possible in order to create the biggest number.

The biggest three digit number that comes to mind is 999. The smallest three-digit that can divide that is 111.

So 999/111 = 9

We classify the types of equ-triangles starting with biggest to smallest.

The biggest is obviously the equ-triangle with a side length of 5, the smallest equ-triangle has a side length of 1.

Count the number of equ-triangles for each class. The first one has already been filled out for you.

Now once you complete the table, just find the total up the number of equ-triangles for the table.

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I believe that  the  largest possible value of M/N  = 9

Thanks! It took me about five minutes of pure guessing-and-checking to get the answer but I guess it was worth it.

Hello Melody,

I understand your confusion; I believe the grid is a three-by-three grid.

The answer is 20160, but I'm not sure why.

Thanks.

If a and b are the roots of x^2 - 4x + 1 = 0, find a^3 + b^3.

Hello Guest!

\(\{a,b \}\subseteq\{x_1,x_2 \}\)

\(x^2 - 4x + 1 = 0\\ x=2\pm \sqrt{4-1}\\ x_1=2+\sqrt{3}\\ x_2=2-\sqrt{3}\)

\(a^3+b^3=(2+\sqrt{3})^3+(2-\sqrt{3})^3=26+15\sqrt{3}+26-15\sqrt{3}\)

\(a^3+b^3=52\)

  !

I understand the second question and redid it myself. But not really the first. So what is the sum of all possible values of the constant k? Clarification would be appreciated. Thank you!!!

I'm sorry about the first question. Yes, "k" is a constant. Thank you for the help. Hope that clears things up!!

Thank you!!!

Sin (A+B) = SInA CosB + sinB Cos A

and

Cos(A+B) = cos A Cos B  -  Sin A Sin B

So the first  1/2 would be

 sqrt3/2 * sqrt2/2  +  sqrt2/2 * 1/2  

        to this you would add the  Cos (A+B) part

sin (60 + 45)   + cos (60 + 45)       =      do you remember angle sum trig identities?

Thank you so much again!!!!

Distance formula between 2 points:

d^2 = (x1-x2)^2 + (y1-y2)^2

100 = (2- -6)^2 + (5 - y2)^2

100 = 64 + (5-y2)^2

36 = (5-y2)^2

6 =+- ( 5- y2)

6 = y2-5             (since we want y2>0)

11 = y2    

Just mutiply every element by 2

This will give you your answer....

y  = 4sin^2(x) -  12sin (x)  +  7

Take the derivative and set to 0

y '=  8sinxcosx   - 12cosx  = 0

So

8sinx cos x  -  12cos x  =  0      factor

4cos (x)  ( 2sin (x)  - 3)  = 0       set each factor to 0  and solve for x  and we have that

x  = pi/2  or 3pi/2    sin x  =  3/2  ( this is not possible )

So....the max occurs at   x  = 0

At  pi/2  the function value is

4sin^2(pi/2) - 12 sin(pi/2)  +  7   =   -1

At (3pi/2)  the function value  is 

4sin^2 (3pi/2) - 12 sin (3pi/2)  +  7  =

4  + 12  +  7   =     

23   =   the max value

14.8 = 16 e^-k(36)       Solve for   k = .002165598

8   = 16 e^{-.002165598 t}      Solve for t = t_1/2 = 320 days

Are your measurements in decimal degrees or Radians?

I personally do not know how to prove it but with all roots there is always 'that' number of roots,

so square root has 2 values

cube roots have 3 values

4th roots have 4 values  etc

I am sure if you google it you will find a proof.

Attn: chenxander

I am still waiting for your response.

thank you so much 

This question is actually very ambiguous.

Sometimes an outcome appears to be valid but if it is rotated it becames invalid.

Since rotations are considered euqivalent then that should mean that the original outcome is also invalid.

I tried to take this into acoount with my answer.  I have said both must be invalid.

It is the diagonal condition that causes this problem.

---------------

Anyway guest, you have not attempted to explain how you got your answer. So it is impossible for anyone to consider your logic.

The terms are

2, 7 , 12, 17

Thanks  answering Guest.

8 students are planning on splitting up into 3 (non-empty) study groups. How many ways can they split up into these study groups?

Note: It is possible that a study group contains only 1 person.

8 Students

If only the number of students in each group counts then

S*S*S*S*S*S*S*S       chose 3 stars from the 7   =  7C2  = 21 ways

(This is the same answer as guest got )

OR

If the exact students in each group is important then:

1,1,6        8C6               = 28

1,2,5         8C5*3           = 168

1,3,4         8C4*4           =  280

2,2,4         8C4*4C2      = 420

2,3,3         8C3*5C3      = 560

28+168+280+420+560 = 1456  ways

The first option looks right to me.  The second one I am much less sure about.

No guarentees for either.

Asker guest:  If you have a number answer can you please share it with the rest of us.

(If you believe it to be wrong please say why you think so.)