All Answers 358317 Answers

it is wrong it came with a hint though "

Find the number of ways to plant quadrants A and C.

Then the number of ways of planting quadrants B and D depends on whether A and C have the same flower or not, so divide into these cases."

Possible  10's digit         Units  digit

           0                               5, 6,7,8,9

           1                               7, 8, 9

           2                               9

We  have  9  possibilities  for  the  hundreds  digit in each case.....so....

9 ( 5  + 3  + 1)   = 

9 (9)  = 

81  such three-digit numbers

Here's  my  take

Start  with  the  top left quadrant -   we  have  4 choices

In the top right quad, we  have  3 choices  (we can't plant  the  same type as in the top left quad)

In the  bottom right quad,  we  again have 3 choices  ( we  can choose to plant the type of flower planted in the top left quad  as well as  the other  2 types ....we   can't plant the  same type as in  the top right quad)

The  sane number of  choices  apply to  the bottom left quad as  with the bottom right quad

So

4  * 3 * 3  * 3   =     108 possible  gardens

P(first card  is a heart  and second  is  a black "number" card)  =

1/4   *   18 / 51   =

1/4  *  6 / 17  = 

6  / 68   =

3 / 34

 2x^2 = -2y^2 + 12x - 4y + 28

To  make  things easier,  divide  through  by 2  and rearrange  as 

x^2  - 6x  +  y^2 + 2y  =  14         complete  the  square on  x and  y

x^2  -  6x  +  9  +  y^2 + 2y  + 1  =   14  + 9  +  1

(x - 3)^2   +  (y + 1)^2  =   24

The  radius  =  sqrt (24)   and  the diameter     =   the side of the  square  =  2sqrt (24)

So.....the   area of the square =  (2sqrt 24)^2   =  4 * 24   =  96  units^2 

Don't blame me that wat the question said

1)  Largest possible degree would be 6  .....it could be LESS

2)  if  b = -1/2  the x⁶ terms would be gone      and the   x²   terms would be gone too.....    degree zero

thats wrong

Not enough info in the question to answer.....

13 / 52 is correct..... then there are 51 cards left of which  18 are black number cards    ace is not a number   ( not 20)

A = 3, B = 1, and C = 9.

this is a website not ELA

127-85 = 42

...Oh yah....   Welcome , DS !

time at 50 mph   =    d / 50

time at  60 mph = d/60

total distance = 2d

distance / time = average

2d / ( d/50 + d/60)

2 / (1/50 + 1/60) = 54.5 m/hr

y=ax³+bx²+cx+d
the constant term is d


if their polynomials have the same constant term  then the product is
(ax³+bx²+cx+d)(rx³+ex²+vx+d)
the only thing we know is the constant terms are the same
we notice that the last term of the product will be the product of the constant terms

so the last term will be d²

the last term is 16
d²=16
sqrt both sides
d=4

the constant term of Jamie's polynomial is 4

$\frac{5850-3350}{3350}$ plug it into your calculator.

WLOG $g\equiv 0$. $3+0=\boxed{3}$.

$(x^4-3x+2)^2$

Upon reduction to $(x^4+x+1)^2$ we see $x^3$ is impossible to get so the answer is $\boxed{0}$.

Clearly $-2-x\ne x^3+2x+2$ for most $x$. Fix the problem please.

You can fill in the dashed boxes. Then solve the sudoku. There are sudoku solvers online if you are lazy :P

Substitute $x=0$ and find what you get. 

Not fucking helpful

Let the constant term of the 3rd degree polynomial equal a. Since the constants in the 3rd degree polynomial multiply to equal the constant in the 6th-degree polynomial, \(a^2=81\). Since none of the signs in the 6th-degree polynomial are negative, the polynomials of Jamie and Billy must both have all negative signs or all positive signs, but since it specified monic polynomials, then it must be that all of them are positive, so the answer is just \(\boxed{9}\)