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I am Lyosha [343]
3 years ago
11

Sophia works in a clothing store and earns $40 per day. She earns an extra $5 for each outfit she sells. If Sophia wants to earn

at least $70 per day, which inequality shows the minimum number of outfits, n, that she should sell? (5 points) 40 + 5n ≤ 70, so n ≤ 6 40 + 5n ≥ 70, so n ≥ 25 40 + 5n ≤ 70, so n ≤ 25 40 + 5n ≥ 70, so n ≥ 6
Mathematics
2 answers:
andrew-mc [135]3 years ago
7 0

Answer:

In order to find this answer, you will need to be able to make an inequality.

Step 1: Sophia makes $40 so start with that. 40

Step 2: Sophia also makes $5 per every outfit that she sells, so add that plus your variable (n). 40+5n

Step 3: Determine if Sophia wants to make more, equal to or more, less, or less than or equal to than what she wants to make per day. 40+5n *draw a greater than or equal to sign.

Step 4: Sophia wants to make more than or equal to what she makes per day. You will need to find what she wants to make more than. In this she wants to make more than or equal to $70.

Equation: 40+5n *greater than or equal to sign* $70

Step-by-step explanation:

oksano4ka [1.4K]3 years ago
3 0

Answer:

  40 + 5n ≥ 70, so n ≥ 6

Step-by-step explanation:

  40 + 5n ≥ 70

           5n ≥ 30

              n ≥ 6

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Step-by-step explanation:

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Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.

We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.

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The amount of fence needed to enclose the external and the internal divisions is

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We know the total fencing is 1/2 miles long, thus

\displaystyle 4x+3y=\frac{1}{2}

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Substituting x

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\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}

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Solving for y

\displaystyle y=\frac{1}{12}

And also

\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}

Compute the second derivative

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Thus, the maximum area is

\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}

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