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tigry1 [53]
3 years ago
5

Plz help! I will make you the brainliest! Thanks!

Mathematics
2 answers:
NeX [460]3 years ago
6 0
-5x < 10

Divide the two sides by -1.
If an inequality form is divided by negative number, the sign will change.
-5x < 10
-5x/-1 > 10/-1
5x > -10

Then divide the two sides by 5
5x/5 > -10/5
x > -2

The answer is -2
Andrews [41]3 years ago
4 0
So to start you will need to divide by -5 then -5 divided by -5 cancelle
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Find the measurements (the length L and the width W) of an inscribed rectangle under the line with the 1st quadrant of the x &am
Leni [432]

The question is incomplete. Here is the complete question.

Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = -\frac{3}{4}x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.

Answer: L = 1; W = 9/4; A = 2.25;

Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:

A = x.y

A = x(-\frac{3}{4}.x + 3)

A = -\frac{3}{4}.x^{2}  + 3x

To maximize, we have to differentiate the equation:

\frac{dA}{dx} = \frac{d}{dx}(-\frac{3}{4}.x^{2}  + 3x)

\frac{dA}{dx} = -3x + 3

The critical point is:

\frac{dA}{dx} = 0

-3x + 3 = 0

x = 1

Substituing:

y = -\frac{3}{4}x + 3

y = -\frac{3}{4}.1 + 3

y = 9/4

So, the measurements are x = L = 1 and y = W = 9/4

The maximum area is:

A = 1 . 9/4

A = 9/4

A = 2.25

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Answer:

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