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exis [7]
3 years ago
15

Grayson charges $35 per hour plus a $35 administration fee for tax preparation. Ian charges $45 per hour plus a $15 administrati

on fee. If h represents the number of hours of tax preparation, for what number of hours does Grayson charge more than Ian?
h less-than 2
h greater-than 2
h less-than 5
h greater-than 5
Mathematics
2 answers:
NikAS [45]3 years ago
8 0

Answer:

h less-than 2

Step-by-step explanation:

frosja888 [35]3 years ago
6 0

Answer:

h less-than 2

Step-by-step explanation:

Let

h ---> the number of hours

y ---> the total charge in dollars

we know that

The linear equation is slope intercept form is equal to

y=mh+b

where

m is the slope or unit rate of the linear equation

b is the y-intercept or initial value of the linear equation

we have

<em>Grayson</em>

The slope is equal to m=\$35\ per\ hour

The y-intercept is b=\$35

so

The total charge is equal to

y=35h+35 ----> equation A

<em>Ian</em>

The slope is equal to m=\$45\ per\ hour

The y-intercept is b=\$15

so

The total charge is equal to

y=45h+15 ----> equation B

For what number of hours does Grayson charge more than Ian?

The inequality that represent this situation is

35h+35 > 45h+15

solve for x

Group terms

35-15 > 45h-35h

20 > 10h

Divide by 10 both sides

2 > h

Rewrite the expression

h< 2\ hours ----> h less than 2 hours

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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:

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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:

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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

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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:

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The money to be charge for each set is:

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

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