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Vlada [557]
3 years ago
13

A restaurant used 15 pounds of ground beef to make hamburgers for the day. How many burgers did they make if each burger is 6 ou

nces? (16 ounces = 1 pound) Explain your answer and show all conversions.NEED ANSWERS TODAY BECAUSE IT DUE TOMORROW!!
Mathematics
2 answers:
Helen [10]3 years ago
8 0

Answer: 40 hamburgers

Step-by-step explanation:

15 pounds x 16 ounces=240 ounces of hamburger meat

Each burger is 6 ounces 240 divided by 6= 40 hamburgers

patriot [66]3 years ago
7 0

They were able to make 40 burgers with it

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If the area of the table is 36 inches by 24 inches, will they have enough tiles to cover the table? If not, how many more will t
Aliun [14]
How many tiles do they have?
36*24=864 If the amount of tiles they have is less than 864, subtract 864 by the amount of tiles.
8 0
3 years ago
A.75/1 dollars per daythe cost $75 for each day.
Rama09 [41]

Answer:

D

Step-by-step explanation:

Count by 25. Its a no-brainer honestly.

5 0
3 years ago
Read 2 more answers
2(x+4)+6=22 what is the answer
mario62 [17]
2(x+4)+6=22
2x+8+6=22
2x+14=22
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2x=8
x=4
5 0
3 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
The farmer needed to fertilize 520 apple trees. He fertilized 20
grigory [225]

Answer: 40 is the answer

Step-by-step explanation:

Farmer fertilized 20  trees per day for 10 days=20*10=200

total trees- fertilized tree=leftover trees

520-200=left over trees

320=left over trees

per day fertilized tree=left over trees /8

per day fertilized trees=320/8

per day fertilized trees=40

If this answer help you mark me brainlist

3 0
3 years ago
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