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The real part is equal to 1/4.

36 bars everyday

in 10 days bars 

received by resturant

  = 10 * 36

  = 360 bars

Let cost of 1 pie be 'x'

So, cost of 1 cake = 2x

     So, 2 cake + 3 pie = $56

           2 (2x) + 3 (x) = 56

             4x + 3x = 56

            7x = 56

              x = 8

 Cost of cake = 2x = 2 * 0

                      = $16

1332

37

1369

1369

37

1332

1369

1, 37, 1369

37x37

1x1369

1369x31

The discriminant of this quadratic is

\(b^2 - 4ac = (-m+2)^2 - 4(m)(m-8) = -3m^2 + 28m + 4\)

For starters, the discriminant has to be nonnegative, which limits our values of m to the solutions of -3m^2 + 28m + 4 >= 0. So, m can be 0, 1, 2, 3, or 4. From there, it's easy to try each individual value until you find an equation with integer roots. 

Let P(x) = x^5 - x^4 + x^3 - px^2 + qx + 14. From the remainder theorem, if P(x) is divisible by x+2, then P(-2) = 0. (The same holds true for x-1.)

So,

P(-2) = -32 - 16 - 8 + 4p - 2q + 14 = 0

P(1) = 1 - 1 + 1 - p + q + 14 = 0

4p - 2q = -42

-p + q = -15

Solving this system of equations will yield the answer.

How do I fing a

As I have already said.

1) I already found c in terms of a and b

2) I have already divided by (x+1)  using algebraic long division, to get the  b in terms of a

3) Now I need to divide that last answer by (x+1)^2 = x^2+2x+1   Using algebraic long division

This will give me a remiander in 'a'.  I know the remainder has to be 0 so put it equal to 0 and solve for a.

Bingo I will have my answer.

I'm not 'essentially' doing anything.

If a polynomial say f(x) is divisable by   (x+k)   then  f(-k) must be zero.     REMAINDER THEOREM

[  f(-k)=the remainder and the remainder in this case will be 0  ]

I have used this fact to help with my solution.

Are you Frosty or are you someone else?  

The complex numbers $z$ and $w$ satisfy $|z| = |w| = 1$ and $zw \ne -1.$

(a) Prove that \(\overline{z} = \frac{1}{z}\quad and \quad \overline{w} = \frac{1}{w}\)

(b) Prove that \(\frac{z + w}{zw + 1}\) is a real number.

\(Let \quad z=e^{i\theta}=cos\theta+ isin \theta\qquad\\ and \quad w=e^{i\alpha}=cos\alpha +isin \alpha\\ \theta+\alpha \ne (2n+1)\pi\\~\\ Prove \;\; \bar z=\frac{1}{z} \\\bar z=cos\theta-isin\theta\\ \frac{1}{z}=(e^{i\theta})^{-1}\\ \frac{1}{z}=e^{-\theta i}\\ \frac{1}{z}=cos(-\theta)+isin(-\theta)\\ \frac{1}{z}=cos(\theta)-isin(\theta)\\ \frac{1}{z}=\bar z \qquad QED \)

Square of 37 = 1369

The complex numbers \(z\) and \(w\) satisfy \(|z| = |w| = 1\) and \(zw \ne -1.\)

(a) Prove that \(\overline{z} = \frac{1}{z}\) and \(\overline{w} = \frac{1}{w}.\)

(b) Prove that \(\frac{z + w}{zw + 1}\) is a real number.

sorry about that :(

still confused, how would you find \(a\)?

so your essentially putting ?\(h(x)=0\)?

Please present you question properly.

Without unrendered LaTex

I don't understand what is being asked for.    

I put a great deal of time into this answer Frosty and I do want a response from you..

670% = 670/100 (/10) = 67/10 = 6 7/10.

Hint:

1)  Use the quadratic formula to find the two solutions to the equation:  7x² - x - 3  =  0.

      Leave them in reduced radical form.

      Label one solution as 'a' and the other as 'b'.

      Note that each solution is in the range:  -1 < x < 1,

2)  1 + a + a² + a³ + ...  is an infinite geometric series, whose sum can be found

     using the formula:  Sum  =  (first term) / [ 1 - (common ratio) ]

     the first term is 1, and the common ratio is the solution that you found in step 1.

     This formula applies because the common ration is in the range -1 < r < 1.

3)  You can find the sum of 1 +  + b² + b³ + ...   in a similar way and multiply these

     two answers.

For each ball, there are three choices of which box to put it in. So, the answer is 4^3 = 64.

This question has now been answered:

https://web2.0calc.com/questions/volume-of-a-cube_1#r1

1. 20, 40, 60, ... , 500, 520

2. 80, 100, ..., 1000

3. 32, 40, ...,  72, 80

4. a) 2, 4, 6, ..., 122, 124

b) 3, 6, 9, ..., 120, 123

c) 6, 12, ..., 114, 120

d) combine the list of numbers in a) and b)

5. 102, 104, ..., 998, 1000

b) 108, 117, ..., 999

c) 108, 126, ..., 990

d) combine the list of numbers in a) and b)

Unfortunately, we can't help you if there is no question posted.