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allsm [11]
2 years ago
9

Can you help me i Don’t understand

Mathematics
1 answer:
snow_tiger [21]2 years ago
7 0

Answer:

  A.  {-1, 2, 7}

Step-by-step explanation:

For a set of ordered pairs, the domain is the set of first-numbers (x). The range is the set of second numbers (y, or g(x)).

The first step here is to find the g(x) values for each of the listed x-values.

The attachment shows a table.

Often, we like to list the domain and/or range values in numerical order, eliminating any duplicates. That is, the set of range values {2, -1, -1, 7} would be listed as ...

  {-1, 2, 7} . . . . . . matches choice A

_____

You find the g(x) values by substituting for x and doing the arithmetic.

  g(-2) = (-2)² -2 = 4 -2 = 2, for example

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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

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Step-by-step explanation:

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