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I agree with your real solutions, Alan

I also thought that the complex solutions to  y^2 - 4y + 7 =  0    were answers

But...Wolframalpha  gives  these  answers for the complex solutions :

https://www.wolframalpha.com/input/?i=y%5E2+-++abs+%284y+-+2+%29++%2B+5+%3D0

Do you see how they got those  ???

yes, it's true

Let's see....    put in 5 for x   and  - 1 for  y    

-1 = 2 (5) - 11     ?    WHat do you think?    True or false?

D. (2, 1)

From the origin....we go 2 to the right on x    and 1 up on y

What is the correct answer  ???

"Solve y^2 - |4y - 2| + 5 = 0."  I'm going to assume only real solutions are required.

(1) Suppose 4y-2 is positive.  Then we have y²-4y+7=0    but b²-4ac = 16 - 28 <0 so solutions are complex.

(2) Suppose 4y-2 is negative.  Then we have y²+4y+3=0. so y = (-4±√(16-12))/2 or y = -1 and y = -3

Check:

If complex solutions are required as well, just solve the quadratic in (1).

Thank you CPhill

C.  3.14 x^2

A. 12.50m + 1.95n

B   5  -  x

A.   6n  - 8

B.  4^3  -  a

B.   4x  - 8  =   112

We have 12 ways to choose the first person in line

11 ways to choose the second

10 to choose the third....etc...

So

12 * 11 * 10 * 9 *.........* 3 * 2 * 1      =   12!

School calculators are sold by one company only in packages of 11 and packages of 15.

What is the largest number of calculators it is possiblee to purchas without opening a package?

Schulrechner werden von einer Firma nur in Paketen von 11 und Paketen von 15 verkauft.
Was ist die größte Anzahl an Taschenrechnern, die Sie kaufen können, ohne ein Paket zu öffnen?

Hello Guest!

The largest number of calculators you can buy without opening a package is

\(=n \times(11+15)\ |n\in \mathbb{N}\ and\ n \nearrow \infty\)

The smallest number of calculators you can buy without opening a package is 11.

  !

1. How many paths of minimum length are there from a to b  in the grid below?

Find the polynomial \(p(x)\), with real coefficients, such that  \(p(x^3) - p(x^3 - 2) = [p(x)]^2 + 12\)
for all real numbers \(x\).

\(p(x) = 6x^3-6\)

I am not sure that it is right but this is my logic.

It is a circle and rotations are considered the same.

In order to deal with this I have fixed on number in place. 

I do not think it makes any difference which initial number you use so the chance of getting this first place right is 1.

Say that first place was filled with an A number. Then the position of the other two As is already fixed. there are 2!=2 ways the other A's can be placed.

There are three fixed places for the Bs  so that is 3! ways and the same for the C's.

so that is where I got  2!3!3! from

There are 9000 four-digit numbers, from 1000 to 9999, inclusive. Rather than counting the numbers with repeated digits, we'll count the numbers without a repeated digit. In this case, there are 9 choices for the first digit (all except 0), 9 choices for the second digit (all except the first), 8 choices for the third digit (two are already picked), and 7 choices for the fourth digit (three are already picked). Therefore, there are 9*9*8*7 numbers without a repeated digit, leaving 9000 - 9*9*8*7 numbers with a repeated digit. To find the percent, we divide this result by 9000, so we get 

\(\frac{9000-9\cdot9\cdot8\cdot7}{9000}=\frac{1000-504}{1000}=.496\)

So, 49.6%

I cannot see a problem with your logic CU.

It would be good if some other knowledgable person also voiced their agreement though.

I am closing this as sinclairdragon is obviously not interested in discussing anything.  

He is only interested in the final answer.

I know this becasue he has reposted.  I have locked that one too.

I do not want this site to be known primarily as a cheat site.

Here is your original post you lazy git.

John has an ample supply of 10-cent stamps and 15-cent stamps. 

He will buy an ample supply of stamps of one more denomination, which is a whole number. 

He needs to be able to combine the stamps to make any value of 30 cents or more. 

In cents, what is the greatest denonmination of stamp he can buy which will enable him to do this?

My attempt:

\(\begin{array}{|rcll|} \hline \text{cent} & \text{any value of } x \text{ cents or more} \\ \hline 2 & x = 14 \\ 3 & x = 18 \\ 4 & x = 22 \\ \color{red}6 & \color{red}x = 30 \\ 7 & x = 34 \\ 8 & x = 38 \\ 9 & x = 42 \\ 11 & x = 90 \\ \hline \end{array} \)

This is a repeat question and as far as I remember you did not make any imput at all last time.

I answered it then, as idid others, and you can comment on the original thread if you want top continue it.

Good work CU         Give yourself a point!

Of course if family members are all differentl then there are   12! ways

That would be good. Thanks Jenny.