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Thanks!

Thanks!

\(10^{\log(x)} = x\\ \text{so yes}\\ 10^{\log(6)} = 6\)

\(\text{one real solution means the discriminant is 0}\\ \text{for the quadratic }ax^2 + b x + c \text{ the discriminant }\\ D = b^2 - 4ac\\ \text{here we have}\\ D = m^2 - 4(m+n) = 0\\ m^2 -4m - 4n = 0\\ m = 2 \left(1\pm\sqrt{n+1}\right)\)

\(\text{we're told that }m \text{ is unique, so }\\ 1+\sqrt{n+1} = 1 - \sqrt{n+1}\\ 2\sqrt{n+1} = 0\\ n=-1\)

I answered the original posting

\(\text{the infinite power tower equals 2, so }\\ \sqrt{x}^{\sqrt{x}^{\sqrt{x}^{\dots}}} = (\sqrt{x})^2 = 2\\ x = 2\)

\(P[x] = \dbinom{7}{x} p^x (1-p)^{7-x} = \\ \dbinom{7}{5} (0.75)^5 (0.25)^2\\ \text{and I leave you to crunch those numbers}\)

We can split this into two parts

1 + 1/6 + 1/36 + ...   and

1/2 + 1/12 + 1/72 + ......

The sum of the two series is

1 /(1 - 1/6)  +   (1/2) / (1 - 1/6)  =

6/5 +  3/5   =

9/5

The common difference, r, = x

So...using the formula fot the sum of an ifinite geomertic series, we have

4 / { 1 - x) > 8    

1 /(1 - x) > 2

1 > 2 ( 1 - x)

1 > 2 - 2x

2x > 1  

x > 1/2

However....x  must also be < 1

So....

1/2 < x < 1  or   ( 1/2, 1 )

Yes. I have been trying to work this. I think the question is missing information. I do not know how to post a diagram.

You can use either number, but at the end you shouldhave done both.

Someone (You?) posted this Q the other day....it was unable to be answered....   Can you post the diagram? 

Last one

x + y + z  = 4           (1) 

5x + 5y + z = 12      (2)

x - 4y + z = 9    ⇒   -x  + 4y - z  = -9     (3)

Add (1) and (3)    and we have

5y  = - 5   

y = -1

Put this into  (1)

x - 1 + z = 4

x + z = 5

z = 5 - x

Using (2)  and making the substitutions we have

5x + 5(-1) + 5 - x = 12

4x  = 12

x = 3

And

z = 5 - 3 = 2

So {x, y, z}   = { 3, -1, 2 }

it sure did help

Second one

x + y -  z = -1             (1) 

4x + 4y - 4z = -2     ⇒    x + y - z = -1/2    (2) 

3x + 2y + z = 0         (3)

Note GM....there are no solutions possible

The reason why???....look at (1) and (2)

x + y - z     cannot  equal  -1    and -1/2   at the same time  !!!!

Whatever x + y - z    is    ....it must lead to the same result every time  !!!

The answer is C) 103

7(6+2)2-3^2

(6+2) = 8

7*8*2-3^2

3^2 = 9

7*8*2-9

7*8*2 = 112

112 - 9

103

Hope this helps ;P

The best way to see this  is to take a "reverse" approach

The total number of possible teams is  C(14, 6)  = 3003 teams

Now assume that the triplets are all on one team....and for the remaining players, we can choose any 3 of the remaining 11...so  this  gives C(11,3) = 165 teams

So...the total number of teams that don't include all three triplets is

3003 - 165 = 

2838 teams

It might be my answer not my question that needs the heart. Idk

C. B+9

Let each square be divided into 4 equal parts....so we have 4 squares * 4 parts each = 16 total parts

In the top left square, using symmetry,  1 of the 4 parts is shaded

And the same for the bottom right square

And for the top right square,  2  of the 4 parts are shaded

And the same for the bottom left square

So...the shaded portion is

[ 1 + 1 + 2 + 2]  / 16 =

6 /16 =

3 / 8

These can be lengthy !!!

x + y + z = 8     (1)

x - y - z = 8        (2)

2x + y + 2z = 16    (3)

Add (1) and (2)    and we get that    2x = 16 ⇒  x = 8

Put this into (2) and (3)  for x

8 - y - z = 8   ⇒      -y - z = 0      (4) 

2(8) + y + 2z =  16   ⇒   16 + y + 2z = 16 ⇒   y + 2z = 0     (5)

Add (4) and (5)  and we get that

z = 0

Using (1)

8 + y + 0 = 8

y = 0

So

{ x, y , z }   = { 8, 0 , 0}