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allsm [11]
3 years ago
13

Will mark brainliest! (problem below)

Mathematics
2 answers:
Mamont248 [21]3 years ago
6 0

Answer:

Helena

Step-by-step explanation:

7y^2z is not modified in either the answer or the question.

6yz^2 - 3yz^2 = 3yz^2

-5 + 2 = - 3

Combining the the partial answers, you get

7y^2 + 3yz^2 - 3

which is the same as the original question.

Vlad [161]3 years ago
4 0

Answer:

Joe

Hope this helps!

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Find the GCF of each expression. Then factor the expression.
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A small business makes cookies and sells them at the farmer's market. The fixed monthly cost for use of a Health Department–appr
DENIUS [597]

Answer:

Cost function:

C(x)=790+2.88x

Revenue function

R(x)=6x

Profit function

P(x)=3.12x-790

Dozens needed for a specific profit P

x=\dfrac{P-790}{3.12}

If 150 dozen cookies are sold the profit is negative, so the business is loosing $322.

Step-by-step explanation:

We can list the costs as:

- Fixed monthly cost, $790/month.

- Variable costs, $0.24/cookie, which are (12*0.24) = $2.88 a dozen.

Then, we can write the cost function C(x) as:

C(x)=790+2.88x

being C(x): the monthly cost and x: the number of dozens produced per month.

The revenue can be calculated as the price ($6 a dozen) multiplied by the number of dozens x:

R(x)=6x

The profit can be calculated substracting the total cost C(x) from the revenue R(x).

P(x)=R(x)-C(x)\\\\P(x)=(6x)-(790+2.88x)=(6-2.88)x-790\\\\P(x)=3.12x-790

The number of cookies (in dozens) that must be produced and sold for a monthly profit can be calculated from the previous equation for P(x).

For a monthly profit P, the number of dozens that need to be sold are:

P=3.12x-790\\\\P+790=3.12x\\\\\\x=\dfrac{P-790}{3.12}

If 150 dozen cookies are sold, the profit made is:

P(150)=3.12(150)-790=468-790=-322

The profit is negative, so the business is loosing $322.

6 0
3 years ago
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