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AlekseyPX
3 years ago
14

The area of the conference table in Mr. Nathan’s office must be no more that 175 ft^2. If the length of the table is 18 ft more

that the width, x, which interval can be the possible widths ?
Mathematics
1 answer:
hodyreva [135]3 years ago
6 0

Answer:

(0, 7]

Step-by-step explanation:

Let's call the width W and the length L.  The length is 18 feet more than the width, so:

L = W + 18

The area can be no more than 175, so:

A ≤ 175

LW ≤ 175

Since L = W + 18:

(W + 18) W ≤ 175

W² + 18W ≤ 175

W² + 18W - 175 ≤ 0

(W - 7) (W + 25) ≤ 0

-25 ≤ W ≤ 7

However, the width of the table can't be nonpositive, so:

0 < W ≤ 7

Or in interval notation, (0, 7].

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A rectangular area is to be enclosed using an existing
Vesnalui [34]

Answer:

\text{Dimensions: 25 x 50},\\\text{Area: }1,250\:\mathrm{m^2}

Step-by-step explanation:

Let the one of the side lengths of the rectangle be x and the other be y.

We can write the following equations, where x will be the side opposite to the wall:

x+2y=100,\\xy=\text{Area}

From the first equation, we can isolate x=100-2y and substitute into the second equation:

(100-2y)y=\text{Area},\\-2y^2+100=\text{Area}

Therefore, the parabola -2y^2+100y denotes the area of this rectangular enclosure. The maximum area possible will occur at the vertex of this parabola.

The x-coordinate of the vertex of a parabola in standard form ax^2+bx+c is given by \frac{-b}{2a}.

Therefore, the vertex is:

\frac{-100}{2(-2)}=\frac{100}{4}=25

Plug in x=25 to the equation to get the y-coordinate:

-2(25^2)+100(25)=\boxed{1,250}

Thus the vertex of the parabola is at (25, 1250). This tells us the following:

  • The maximum area occurs when one side (y) of the rectangle is equal to 25
  • The maximum area of the enclosure is 1,250 square meters
  • The other dimension, from x+2y=100, must be 50

And therefore, the desired answers are:

\text{Dimensions: 25 x 50},\\\text{Area: }1,250\:\mathrm{m^2}

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3 years ago
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Answer:

A) 0 and 13

Step-by-step explanation:

64 - 37 = 27

27/2 = 13.5

13 teachers can teach 3 classes at max

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