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a+b = 1      square both sides 

(a+b)^2 = 1 

a^2 + b^2 + 2 ab = 1           but we are given    a^2 + b^2 = 2     substitute this in 

        2       + 2 ab = 1 

               ab = -1/2 

Now 

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

                       a^3 + b^3    + 3ab (a+b)                                 Substitute in ab = -1/2   and a+b = 1 

                        a^3 + b^3   + 3 (-1/2) (1) 

(a+b)^3 =                    a^3 + b^3    - 3/2                                  if a+b = 1    then (a+b)^3 = 1     susttiute this in on the left

      1      =    a^3 + b^3 - 3/2                                          add 3/2 to both sides

      5/2  = a^3 + b^3 

I think the 'X' is the multiplication symbol.....

(1+r) c   +   r(1+r)c          =  c ( 1+ r)^2       ??         Expand LEFT side 

c + r c   +   r c  + c r^2    =   c ( 1+r)^2                    Simplify LEFT side a bit

c + 2rc + cr^2                 = c (1+r)^2                       Factor out 'c' on the LEFT 

c ( 1 + 2r + r^2 )             =  c ( 1+r)^2                     Reduce LEFT side 

c (1+r)^2                        =   c (1+r)^2            <=====  TRUE ! !  

\(500,25, 5/4, 5/80\)

But 25 is negative because of the negative sign on the common ratio, so heres the updated list greater than 1.

\(500,-25,5/4\)

There are only 2 numbers of the sequence greater than 1

15 members = 45 minutes

5 members = 135 minutes

20 members = 135/4 minutes = 33.75 minutes

Im gonna hard bash it. Its requires extreme patient which I do not have. I shall be using a calculator to shorten the process.

Oh well....

\(N = 2^{20}+2^{19}+2^{18}+2^{17}+2^{14}+2^{13}+2^{12}+2^{10}+2^{6}+A\cdot2^{5}+B\cdot2^{4}+C\cdot2^{3}+2^{1}\)

\(N=1995842+A\cdot2^{5}+B\cdot2^{4}+C\cdot2^{3}\)

\(1995842 \cong 2\mod7\)

\(A\cdot2^{5}+B\cdot2^{4}+C\cdot2^{3}\cong 5\mod7\)

\(32\cong4\mod7,16\cong2\mod7,8\cong1\mod7\)

\(32+8\cong4+1=5\mod7\)

\(A=1,B=0,C=1\)

\(x^2 + 15x + k = 0\)

\(\Delta = \sqrt{225-4k} = 0\)

\(225-4k = 0\)

\(4k = 225\)

\(k = \frac{225}{4}\)

hmm. are you sure this is it? or can you give some context. Im not sure what the x is doing in this equation, and it doesn't cancel out.

The left hand side would be equal to ^{\([C_0(1+r)](rx+1)\), so maybe \(rx+1=r+1\), so \(x=1\). Thats the only way i can think of}

One ordered pair $(a,b)$ satisfies the two equations $ab^4 = 48$ and $ab^3 = 24$. What is the value of b in this ordered pair?   

                                                     ab⁴  =  48       

                                                     ab³  =  24    

ab⁴ = 48 and ab³ = 24 so let's   

divide both sides of the first by     

respective equal values of the     

second to obtain                              b = 2     

the problem doesn't ask for     

the value of a, but it's                       a = 3      

_.    

                                     x^2 - 5x + 4 = -x^2 - 25x - 14    

combine like terms        2x² + 20x +18  =  0    

divide by 2                      x² + 10x + 9  =  0    

factor                              (x + 1)(x + 9)  =  0    

                                                 (x + 1)  =  0      ==>>  x = –1    

                                                 (x + 9)  =  0      ==>>  x = –9    

                                                         x  =  –1, –9   

_.   

A school orders $99$ textbooks, all for the same price.   

When the bill for the total order comes, the first and last   

digits are obscured. What are the missing digits?   
$_5801.3_   

Consider that the price per book is multiplied by 99.   

And, multiplying by 99 incorporates multiplying by 9.    

The digits of the product of a number multiplied by 9    

will always sum to a total that is evenly divisible by 9.    

5 + 8 + 0 + 1 + 3 = 17  so the sum of the two missing     

numbers will have to add 1 or 10 or 19, etc. to that 17.     

Using brute force, I reckoned that 15801.39 is evenly

divided by 99.  So the missing numbers are 1 and 9.    

And the cost of each textbook was 159.61.    

_.    

\(244224 \pmod {64} \equiv 0 \)

If angle XWZ = 60°, angle WXY = 180° - 60° = 120° because XY and WZ are parallels.

However, this means that Angle ZXW must be less than 120°.

In the problem though, it clearly defines ZXW as 120, meaning this problem is invalid. 

Thanks! :)

No answer. 

In order not to have two yellows next to each other,   

the three yellows must occupy squares 1, 3, and 5.  

That leaves squares 2 and 4 to fill, and two colors   

to do it with. There are only four ways.  

Blue Blue     Red Red     Blue Red     Red Blue   

So, the answer is four ways.    

_.   

Kwan reads a book that is 400 pages long at a rate of 1 page per minute. He reads for half an hour each day, which equals 30 minutes per day.

### Step 1: Calculate the number of pages Kwan reads per day

Since Kwan reads 1 page per minute and reads for 30 minutes each day:

\[
\text{Pages per day} = 30 \times 1 = 30 \text{ pages}
\]

### Step 2: Calculate the number of days required to finish the book

To find the number of days required to finish the book, divide the total number of pages by the number of pages he reads per day:

\[
\text{Number of days} = \frac{400 \text{ pages}}{30 \text{ pages/day}} = \frac{400}{30} \approx 13.33 \text{ days}
\]

Since Kwan cannot finish the book in a fraction of a day, we round up to the nearest whole number:

\[
\text{Number of days} = 14 \text{ days}
\]

### Step 3: Determine the day of the week

Kwan begins reading on a Saturday. We need to find out what day it will be 14 days later.

Starting from Saturday:

- 7 days later: Saturday (end of 1st week)

- 14 days later: Saturday (end of 2nd week)

Thus, Kwan will finish the book on a **Saturday**.

### Final Answer:

Kwan will finish the book on \( \boxed{\text{Saturday}} \).

The answer is \(0\)