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ikadub [295]
3 years ago
7

You have eleven books but only have room for six of them on a shelf. How many possible ways can the books be arranged?

Mathematics
1 answer:
Verizon [17]3 years ago
8 0

Answer:

I would say 720

Step-by-step explanation:

one possibility that comes to my head is stack them on top of each other in many ways

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Im going to middel school and i want to pick the best sport (for girls)
Vadim26 [7]

Answer:

Basketball all the way!

Step-by-step explanation:

3 0
3 years ago
Let f(x)=3x+5 and g(x)=x^2 find (f-g)(x)
dlinn [17]

Answer:

(f - g)(x) = -x² + 3x + 5

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties

<u>Algebra I</u>

  • Function Notation
  • Combining Like Terms

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = 3x + 5

g(x) = x²

(f - g)(x) is f(x) - g(x)

<u>Step 2: Find (f - g)(x)</u>

  1. Substitute:                   (f - g)(x) = 3x + 5 - x²
  2. Rewrite:                       (f - g)(x) = -x² + 3x + 5
7 0
2 years ago
Who ever helps get branliest.​
andrew11 [14]

Answer:

c ?

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Explain answer !! <br><br> ( WILL GIVE BRAINLEST)
Ganezh [65]

Answer:find the solution to the system of equations below.start by adding the equation

{-2x+2y=-8

{2x+y=5

Step-by-step explanation:The given system of equation are:

1----  y= x -4

2----- y=4 x+2

Equation (1) - Equation (2)

→0= x -4 - (4 x+2)

→0=  -3 x -6-----[Adding and subtracting like terms]

Adding 3 x on both sides

→3 x= -6

Dividing by 3 on both sides

→x = -2

→Substituting the value of x in equation 1, we get

y= -2 -4

y =-6

→Solution set =(x,y)=(-2, -6)

Don't used the graph to find the solution.

4 0
2 years ago
A company manufactures and sells x television sets per month. The monthly cost and​ price-demand equations are ​C(x)=74,000+80x
SOVA2 [1]

Answer:

a) $675000

b) $289000 profit,3300 set, $190 per set

c) 3225 set, $272687.5 profit, $192.5 per set

Step-by-step explanation:

a) Revenue R(x) = xp(x) = x(300 - x/30) = 300x - x²/30

The maximum revenue is at R'(x) =0

R'(x) = 300 - 2x/30 = 300 - x/15

But we need to compute R'(x) = 0:

300 - x/15 = 0

x/15 = 300

x = 4500

Also the second derivative of R(x) is given as:

R"(x) = -1/15 < 0 This means that the maximum revenue is at x = 4500. Hence:

R(4500) = 300 (4500) - (4500)²/30 = $675000  

B) Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 80x) = -x²/30 + 300x - 80x - 74000

P(x) = -x²/30 + 220x - 74000

The maximum revenue is at P'(x) =0

P'(x) = - 2x/30 + 220= -x/15 + 220

But we need to compute P'(x) = 0:

-x/15 + 220 = 0

x/15 = 220

x = 3300

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3300. Hence:

P(3300) =  -(3300)²/30 + 220(3300) - 74000 = $289000  

The price for each set is:

p(3300) = 300 -3300/30 = $190 per set

c) The new cost is:

C(x) = 74000 + 80x + 5x = 74000 + 85x

Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 85x) = -x²/30 + 300x - 85x - 74000

P(x) = -x²/30 + 215x - 74000

The maximum revenue is at P'(x) =0

P'(x) = - 2x/30 + 215= -x/15 + 215

But we need to compute P'(x) = 0:

-x/15 + 215 = 0

x/15 = 215

x = 3225

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3225. Hence:

P(3225) =  -(3225)²/30 + 215(3225) - 74000 = $272687.5

The money to be charge for each set is:

p(x) = 300 - 3225/30 = $192.5 per set

When taxed $5, the maximum profit is $272687.5

3 0
3 years ago
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