What are the zeros of x(x + 2)(x

A theater company with 9 actors is putting on a play that has 3 roles. In how many different ways can the director assign actors

Cerrena [4.2K]

He has 9 choices for the first role, and that leaves him 8 people to choose for the second so he has 9*8 = 72 choices for the first two roles; that leaves 7 possible choices for the third role so that gives him 9*8*7 choices or 504 different ways he can assign the roles. i think this is correct.

5 0

3 years ago

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Solve the following System of Three Equations: x−3y+z=−15 2x+y−z=−2 x+y+2z=1

jeyben [28]

Answer:

x = -3 , y = 4 , z = 0

Step-by-step explanation:

Solve the following system:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for z:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Solve for z.

Subtract x - 3 y from both sides:

{z = 3 y + (-x - 15)

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Perform a substitution.

Substitute z = -15 - x + 3 y into the second and third equations:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

x + y + 2 (-15 - x + 3 y) = 1

Hint: | Expand the left hand side of the equation x + y + 2 (-15 - x + 3 y) = 1.

x + y + 2 (-15 - x + 3 y) = x + y + (-30 - 2 x + 6 y) = -30 - x + 7 y:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for x:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Isolate terms with x to the left hand side.

Subtract 15 - 2 y from both sides:

{z = -15 - x + 3 y

3 x = 2 y - 17

-30 - x + 7 y = 1

Hint: | Solve for x.

Divide both sides by 3:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

-30 - x + 7 y = 1

Hint: | Perform a substitution.

Substitute x = (2 y)/3 - 17/3 into the third equation:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Choose an equation and a variable to solve for.

In the third equation, look to solve for y:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Isolate terms with y to the left hand side.

Add 73/3 to both sides:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 = 76/3

Hint: | Solve for y.

Multiply both sides by 3/19:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

y = 4

Hint: | Perform a back substitution.

Substitute y = 4 into the first and second equations:

{z = -x - 3

x = -3

y = 4

Hint: | Perform a back substitution.

Substitute x = -3 into the first equation:

{z = 0

x = -3

y = 4

Hint: | Sort results.

Collect results in alphabetical order:

Answer: {x = -3 , y = 4 , z = 0

6 0

3 years ago

Find the first, fourth, and eighth terms of the sequence. A(n) = −2 ∙ 2x − 1

Nutka1998 [239]

$A(n)=-2\cdot2^{n-1}\\\\\text{Find the first, fourth, and eighth terms}\\\text{It means}\ A(1),\ A(4)\ \text{and}\ A(8).\\\\\text{Substitute}\ n=1,\ n=4\ \text{and}\ n=8\ \text{in}\ A(n):\\\\A(1)=-2\cdot2^{1-1}=-2\cdot2^0=-2\cdot1=-2\\\\A(4)=-2\cdot2^{4-1}=-2\cdot2^3=-2\cdot8=-16\\\\A(8)=-2\cdot2^{8-1}=-2\cdot2^7=-2\cdot128=-256\\\\\boxed{Answer:\ -2,\ -16,\ -256}$

5 0

3 years ago

Which sequence has a common ratio of 2? a{20, 40, 80, 160, 320, 640, …} b{20, 10, 5, 2.5, 1.25, 0.625, …} c{20, 15, 10, 5, 0, -5

Elis [28]

Answer:

A

Step-by-step explanation:

40/20=2

80/40=2

Therefore the common ration is 2

4 0

3 years ago

Question 1,2, and 3 how do i factor those? Can you show the work and explain how?

DiKsa [7]

1: $3n^{2}+9n+6$

notice that each part is divisible by 3

$3n^{2}$ ÷ 3 = $n^{2}$

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes $3(n^{2} +3n+2)$

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

$3(n^{2} +2n+n+2)$

Now that it's rewritten, you can factor out n + 2 from the equation.

the answer is

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is $n^{2} +2n+n+2$ and then each of those by 3, which is $3n^{2} +6n+3n+6$ or $3n^{2}+9n+6$ , our origional equation

2: $28+x^{2} -11x$

So I rewrote this as $x^{2} -11x+28$ (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

(remember, the two numbers should multiply to equal 28, which is our constant.)

$x^{2} -4x-7x+28$

now we can factor out x from the first expression and -7 from the second

$x(x-4)-7(x-4)$

and lastly you factor out x-4,

which would give you

(x-4)(x-7)

Make sure to check your work and make sure it multiplies to $x^{2} -11x+28$

3: $9x^{2} -12x+4$

The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12

So if you rewrite 9 as $3^{2}$ and 4 as $2^{2}$ , the equation becomes