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Gemiola [76]
2 years ago
5

iris charges $50 for her consulting services plus $60 for each hour she works. on one job, iris charged $470. for how many hours

did iris work on this job​
Mathematics
2 answers:
Art [367]2 years ago
8 0

Answer:

the answer is 7 hope I could

Ber [7]2 years ago
3 0

Answer: 7 hours.

Step-by-step explanation:

Take the original amount (470) and subtract the $50 consulting fee. then divide the remaining amount by 60, as that is what she charges per hour. See the work done here:

470 - 50 = 420

420/60 = 7

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ahrayia [7]

Answer:

x = 16

Step-by-step explanation:

Supplementary angles = 180

K + L = 180

137 + 3x - 5 = 180

132 + 3x = 180

3x = 180 - 132

3x = 48

x = 48/3

x = 16

8 0
3 years ago
Can someone help me
Stolb23 [73]

Answer:

I think it is 249 not sure bc i dont know the multiple answers choices

Step-by-step explanation:

5 0
2 years ago
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The table and graph below show the amount of money Mi-Ling and Daniel
Over [174]
Is there a math question and can you show it please
6 0
3 years ago
What is an equation of the line that passes through the points (1,4) and (-2,-8)?
natita [175]

Answer: I know it 4 but I don’t know the equation for it, I can only help you this far.(hope it help)

Step-by-step explanation:

(1,4) and (-2,-8)

(x is the first number of the point which is 1 and -2, y is the second number of the point which is 4 and -8.)

(x1 is 1, it alway the first point they tell you on the problem like (1,-4) is on the problem first so it gonna be x1 and y1, for x2 it -2, it alway the second point in the problem like (-2,-8) is the second point they tell you so it gonna be y1 and y2.)

Do x2 - x1 and y2 - y1(y/x)

-8 - 4/-2 - 1 = -12/-3

-12/-3 divide top and bottom by -3, that would equal to 4

8 0
2 years ago
A company manufactures and sells x television sets per month. The monthly cost and​ price-demand equations are ​C(x)=72,000+60x
myrzilka [38]

Answer:

Step-by-step explanation:

Given

Cost Price c(x)=72000+60x

Price p(x)=300-\frac{x}{20}

Revenue generated R(x)=P(x)\times x

where x=no of units

R(x)=300x-\frac{x^2}{20}

To get maxima and minima differentiate R(x)

\frac{\mathrm{d} R(x)}{\mathrm{d} x}=0

\frac{\mathrm{d} R(x)}{\mathrm{d} x}=300-2\times \frac{x}{20}=0

300=2\times \frac{x}{20}

x=3000

maximum Revenue R(x)=(300-\frac{300}{20})\times 300=4,50,000

(b)Profit=Revenue - cost

Profit=xp(x)-c(x)

Profit=300x-\frac{x^2}{20}-72000-60x

Profit(z)=240x-\frac{x^2}{20}-72000

differentiate Profit to get maximum value

\frac{\mathrm{d} z}{\mathrm{d} x}=240-2\times \frac{x}{20}

x=2400

maximum Profit z=2,16,000

(c)Now company decided to tax the company $ 55 for each set

Profit (z_1)=xp(x)-c(x)-55x

z_1=300x-\frac{x^2}{20}-72000x-60x^2-55x

z_1=185x-\frac{x^2}{20}-72,000

differentiate Profit to get maximum value

\frac{\mathrm{d} z_1}{\mathrm{d} x}=0

\frac{\mathrm{d} z_1}{\mathrm{d} x}=185-\frac{2x}{20}=0

x=1850

P(z_1\ at\ x=1850)=99125

company should charge 207.5 $ for each set

         

6 0
3 years ago
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