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The probability is 5/72.

The answer is 48.

angle B = 120 degrees.

The value of k is 7/2.

There are 12 x 12 = 144 ways to roll the 2 numbers. 

Let's count the amount of ways as shown: 

max digit = 12: (12, 5 - 12) = 8 ways 

max digit = 11: (11, 6 - 11) = 6 ways

max digit = 10: (10, 7 - 10) = 4 ways 

max digit = 9: (10, 8 - 9) = 2 ways 

There are 8+6+4+2 = 20 ways, which makes for a probability of \({20 \over 144} = \color{brown}\boxed{ 5 \over 36 } \)

I'm not sure what this question is asking for.

Do not post A o P S homework.

If a,b,c are the roots of : \(2x^3-6x^2-15x-3=0\)

Then, by Vietas' formulae:

\(a+b+c=3\) 

\(ab+ac+bc=-\frac{15}{2}\)

\(abc=\frac{3}{2}\)

We want: \(a^2+b^2+c^2\)

Using the following identity:

\(a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)\)

Then:

\(a^2+b^2+c^2=3^2-2(-\frac{15}{2})=9+15=24\)

We can rewrite the system of equations by replacing \(x\) with \(-2\) to get

\(y-6=a\)

\(5y-10=2a\).

Multiplying the first equation by \(2\), we have \(2y-12=2a\).

Subtracting this from the second equation says that \(3y+2=0\), so therefore \(y=-2/3\), and so \(a=-20/3\).

n=1;a=400;s=0;m=n;p=0;n=m;cycle:s=s+(n%10);p=p+1;n=(n div 10);if(n!=0, goto cycle,0);m=m+1;if(m<=a, goto5,0);print"Total Sum =",s;print"Total Num =",p

Total Sum = 4204
Total Num = 1092

t==2

LCM[12, 2]^3 ==12^3 ==1,728 ==[6 * 2]^3 ==12^3 ==1,728

Area of triangle DPE  can be found as

Semiperimemter, S,   =   [2.5 + 2 + 1.5 ]  / 2  =  3

Area  =  sqrt [  3 ( 3 - 2) (3 - 2.5) ( 3 -1.5) ] =   1.5  

By the property  of  medians

PC =  2 * PE  =  2 * 1.5  =  3

PA =  2 *  PD =  2 * 2  =  4

And triangles  DPE and APC are similar  with each side of APC twice as long as that of DPC

So scale factor of triangle   APC   to  triangle DPE   =   2

Area of triangle  APC =  area of DPE * scale factor^2  =   1.5 * 2^2  =  6

And by another median property areas of triangle  AEP  plus triangle  CDP   = area of triangle APC

So....area of  AEDC  =   1.5 + 6  + 6   =     13.5

Any x that makes the denominator =  0    is not in the domain

So

x = 9    and x =  -9  are not in the domain

Domain  = 

(-inf , -9) U (-9, 9)  U  (9 , inf)

See the image here

Area of ABC  = 

[DFEC ] -  [DBA] - [AEC]  - [BFC]  =

[6*9 ] -  [ (1/2)(4)(3) ]  - [(1/2) (6) (6) ] - [(1/2) (2) (9) ] = 

54  - 6    - 18  -  9   =

21

Clearly, this is A o P S homework.  Do not post A o P S homework.

If x  = 3  then

3(3)^2  + 2   =   29

So  we need to have

a(3)^2 + 2  =  29

9a  = 27

a = 3

See the graph here : https://www.desmos.com/calculator/srad0t3mxn

3x^2  + 4x  +  1  =  a( x + h)^2 + k

3x^2 + 4x + 1  =  a (x^2 + 2xh + h^2) + k

3x^2 + 4x + 1  =  ax^2 + 2ahx + ah^2  + k

a = 3

4  = 2ah

4 = 2(3)h

4 = 6h

h = 4/6 = 2/3

ah^2 + k =  1

3(2/3)^2 + k  = 1

(4/3) + k =  1

k =  1  - 4/3

k = -1/3

a)  Basketball only =  (1/5) (200 )  = 40

b)   1/5   = 20%       1/4  = 25%

So   100  -  20   -20   - 25  - 13  =   22%  like Football only

The number who like Football   =   (22 + 13) (200)  =  (35)(200)   = 70

9x^2 + sx  +  4                 

Take the positive root of  9x^2 and  both roots of 4   = 3x  and   +/- 2

(3x  + 2)^2   =

9x^2  + 12x +  4

s = 12

Also

(3x -2)^2  =  

9x^2  - 12x + 4

s =  -12

If we stated with 900 and each contained an equal number at the end, then each box contained 300 at the end

Let a,b,c  be the number in each box to start with

b started with b

And after the transfers

b + 18  - 5  = 300

b + 13  = 300

b = 300 - 13  =  287

c started  with     900 - (a + b)

And after the transfers 

(2/3) [ 900 -(a + b) + 5 ]  = 300

905 - a - b =  450

905 - a - 287  =  450

618 - a = 450

a = 618 - 450  =  168

c =  900 - 168 - 287 = 445

So

a started with 168

b started with  287

c started with  445

Suppose james bought x cans of sardines & y cans of tuna

x/y = 4/5

∴  x = 4/5y -- (1)

A total of 22 cans of sardines & y/2 cans of tuna were used

∴ (x - 22)/y/2 = 1/2

∴ x - 22 = y/4 -- (2)

from (1) & (2)

4/5y - 22 = y/4

∴ 4y/5 - y/4 = 22

∴ 16y - 5y = 22 * 20

∴ y = 40

∴ x = 4/5 * 40

∴ x = 32 cans

Let Ben begin with N sweets

Then Calvin had  4N sweets

And Andy had   840  - N  - 4N  =  840   - 5N sweets

When Andy and Ben each give away (1/2)  of theirs we have

N / 2  +  (840 - 5N)/2 +  4N =   620

N / 2  + 420 - (5/2)N + 4N  =  620

420 + 2N  =  620

2N  = 620  - 420

2N =  200

N = 100 =  the number that Ben had at first

1.5 or 1

B = 11
D = 9

In triangle AIB ,

(1/2)A  + (1/2)B  + angle AIB  =180

(1/2) angleA + (1/2) angleB  + 103  = 180

(1/2)angleA + (1/2) angleB =  180  -103

(1/2) ( angleA + angle B)  = 77       multiply through by 2

angleA + angle B   =  154

Angle C =   180  -  154  =   26°