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10 and 4 

x - y = 6     so    y = x-6

x ( x-6) = 40 

x^2 - 6x -40 = 0 

(x-10)(x + 4) =0       shows the positive answer for x = 10    then  the other is 10 - 6 = 4 

Let c be the number of students in the cooking club and d be the number of students in the drama club but not the cooking club. Then we have the following equations:

c = 2d a = c + d

Solving for d, we get d=a/3. Therefore, the number of students who are in the cooking club or the drama club but not both is c−d = 2d−d = a/3.

Since there must be at least three red squares, the red squares must be in squares 1, 3, and 5. That leaves two squares to contend with, squares 2 and 4. We can color those two squares in 4 ways:

Red Yellow Red Blue Red

Red Blue Red Yellow Red

Red Yellow Red Yellow Red

Red Blue Red Blue Red

So the answer is 4​.

....or as a simple ratio

     Edith can run 19   in the time it takes Foley to run 18  

     18 is to 19    as   x is to 24 

18 / 19 = x / 24 

x = 24 * 18/19 = 22  14/19     or     1  5/19  km  behind Edith 

x = the number 

x^2 = 4x - 3      re-arrange

x^2 - 4x + 3 = 0     factor   ( or use Quadrtatic Formula) 

(x-1)(x-3) = 0       shows x = 1  or 3  

x = time of D finish 

E rate = 19/x        time needed for E to finish =  d/ rate = 5 km / ( 19/x) =   5x / 19  

F rate = 18/x

  Howo far does F run in   5x/19 ?      d = r * time 

                                                           d =  18/x  * 5x /19 =  90/19 km 

                                                                   added to his 18 km = 22  14/19 km   ....which is 1 5/19 km behind Edith as she finishes the 24 km 

First, x^2+y^2=14x+48y-8x-26y+16   rearrange as x^2 -6x + y^2 -22y  = 16 

 complete the square on x, y

x^2  - 6x + 9  + y^2 -22y + 121  =  16 + 9 + 121

(x - 3)^2  + (y -11)^2 = 146

This is a  circle centered at (3, 11)  with a radius  of sqrt (146)

The minimum value of y =   11 - sqrt (146) ≈   -1.083

I read the other answers already provided for this particular problem, and I could not make sense of them, so I will post my own approach to this problem. Before we start with the nitty-gritty details of this problem, we should first determine what possible values of A exist in base 63. In the base 63 system, the smallest possible digit is the digit representing 0 in base 10, and the largest possible digit represents 62 in base 10. Therefore, we conclude that \(0 \leq A \leq 62\). This will become important when we generate values of A.

First, I will expand the base 63 number into its more familiar base 10 form by expansion. However, since we only care about the unit's digit of this particular number, I will simplify as much as possible to make the computation easier.

\(A7894321_{63} = 1 \times 63^0 + 2 \times 63^1 + 3 \times 63^2 + 4 \times 63^3 + 9 \times 63^4 + 8 \times 63^5 + 7 \times 63^6 + A \times 63^7\)

The first observation I made is that the unit's digit of \(63^p\) and \(3^p\) for nonnegative integer powers p have the same unit's digit. This occurs because \(63^p \pmod{10} \equiv 3^p \pmod{10}\). This means I can simplify the nasty expansion to the following. In order to ensure that I am writing mathematically, I will define a new function \(U(x)\) which returns the unit's digit of an integer x.

\(U(A7894321_{63}) = 1 \times 3^0 + 2 \times 3^1 + 3 \times 3^2 + 4 \times 3^3 + 9 \times 3^4 + 8 \times 3^5 + 7 \times 3^6 + A \times 3^7\)

I began exploring the powers of 3, and I immediately noticed a pattern among the powers of 3 and their units digit.

Observing the first few values of \(3^p\) should convince yourself that a never-ending pattern exists where 3 raised to the power of 0, 1, 2, and 3 have unique unit's digits, but the pattern repeats thereafter. This essentially means that we can simplify the powers further by applying the rule that \(U\left(3^p\right) = U\left(3^{p \pmod{4}}\right)\)

\(U(A7894321) = 1 \times 3^0 + 2 \times 3^1 + 3 \times 3^2 + 4 \times 3^3 + 9 \times 3^0 + 8 \times 3^1 + 7 \times 3^2 + A \times 3^3\)

Given our observations from the previous investigation of the powers of 3, we can reference that table to determine the unit's digit of the powers of 3.

\(U(A7894321) = 1 \times 1 + 2 \times 3 + 3 \times 9 + 4 \times 7 + 9 \times 1 + 8 \times 3 + 7 \times 9 + A \times 7\)

We can do these multiplications without a calculator unlike before where the calculation would have been unfeasible.

\(U(A7894321) = 1 + 6 +27 + 28 + 9 + 24 + 63 + 7A\)

We could do the addition now, but we can reduce the two-digit numbers to its final digit and do the addition because we only care about the unit's digit.

\(\begin{align*} U(A7894321) &= 1 + 6 + 7 + 8 + 9 + 4 + 3 + 7A \\ &= 38 + 7A \\ &= 8 + 7A \end{align*} \)

We finally have a nice and compact representation of the unit's digit. Of course, we want the unit's digit of this base 63 number to be zero. In order to make \(U(8 + 7A) = 0\). The first part of the addition has a unit's digit of 8. In order to make the sum have a final digit of 0, it would be required that \(U(7A) = 2\).

I will once again try to generate a table of values and observe a pattern

Based on these observations, \(U(7A) = 2\) when \(U(A) = 6\). In other words, the unit's digit of A needs to be 6. Now, all we need to do is list the values of A with a unit's digit of 6. Also, recall from the beginning that we determined that \(0 \leq A \leq 62\).

This means that \(A = \{6, 16, 26, 36, 46, 56\}\). All values in this set are possible values for A.

The equation x^2 + 6x - 5 = 0 is a quadratic equation, and you can solve it for x using the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this equation:

a is the coefficient of the x^2 term, which is 1 in this case.

b is the coefficient of the x term, which is 6 in this case.

c is the constant term, which is -5 in this case.

Now, plug these values into the quadratic formula:

x = (-6 ± √(6² - 4(1)(-5))) / (2(1))

x = (-6 ± √(36 + 20)) / 2

x = (-6 ± √56) / 2

x = (-6 ± 2√14) / 2

Now, simplify by dividing both terms in the numerator by 2:

x = -3 ± √14

So, the solutions to the equation x^2 + 6x - 5 = 0 are:

x₁ = -3 + √14 x₂ = -3 - √14

Seven people must shake hands with six other people.... you might think that is 42 handshakes ....BUT...each handshake involves TWO people so the total number of handshakes is 1/2 of 42 = 21 handshakes. 

There are 4 ducks

One 

When calculating the number of objects that are in pairs, the correct answer is one. For instance, if there are 11 pairs of socks and each pair has 2 pairs of shoes, then the result of adding 2 to 11 is 22. This is equal to 1 pair plus 1 sock. Similarly, adding 2 to 11 gives 22 pairs of shoes, which is 1 pair plus 1 shoe. In both instances, the result of 2 to 11 is 1 unit, even if the total of the items is greater than 1.

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We can write out the base 63 number A7894321(63) in base 10. This gives us the following number:

A * 63^6 + 7 * 63^5 + 8 * 63^4 + 9 * 63^3 + 4 * 63^2 + 3 * 63 + 2 * 63 + 1

We want this number to end with a zero. This means that the units digit of the number must be 0. The units digit of the number is equal to 1 * 1 + 0 = 1. Therefore, A must be a digit that, when multiplied by 1, gives a units digit of 0. The only digits that satisfy this condition are 1, 3, 5, 7, and 9.

Therefore, the possible values of A are 1, 3, 5, 7, and 9

If I understand your question, then:

When A==1 - THE DECIMAL NO =  35896572549181
When A==2 - THE DECIMAL NO =  8323708075012
When A==3 - THE DECIMAL NO =  12262688714179
When A==4 - THE DECIMAL NO =  16201669353346
When A==5 - THE DECIMAL NO =  20140649992513
When A==6 - THE DECIMAL NO =  24079630631680
When A==7 - THE DECIMAL NO =  28018611270847
When A==8 - THE DECIMAL NO =  31957591910014
When A==9 - THE DECIMAL NO =  35896572549181

As you can see, only when A==6, the decimal equivalent ends in a zero "0".

The triangle formed will be a right  triangle with the right angle at  (-1.5, .5 )

In this case, the  circle passing through all three vertices will  lie  at the  midpoint of (2,4)  and (2,-3)   =  (2, .5)

x^2 + bx -28 = 4x + 14    Simplify to 

x^2 + (b-4) x - 42 = 0             If - 7 is a root, puting in -7 for 'x' will result in this equation = 0 

(-7)^2 + (b-4)(-7) - 42 = 0 

49     - 7b + 28 - 42  = 0 

35 - 7b = 0

b = 5 

This is an  isosceles  triangle with equal sides  of 5 and a  base of  6

The circumradius =    abc  /  [ 4 * area ]     where a,b,c are the sides  of the triangle

abc =  5*5*6   =  150

Semi-perimeter=  [  5 + 5 + 6] / 2 =   8    =  s   

Area =  sqrt  [ s (s - a) ( s -b) ( s-c) ]  =  sqrt [ 8 (8-5) (8-5) (8-6) ] =  sqrt [ 144] = 12

4 * area =   48

circumradius=  150 /  48  =  50 / 16  = 25 / 8  =   3.125  

hey, what does the answer say and the explination? it shows as some really weird symbols.