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Neporo4naja [7]
3 years ago
6

A rain gutter is made from sheets of aluminum that are 16 inches wide by turning up the edges to form right angles. Determine th

e depth of the gutter that will maximize its​ cross- sectional area and allow the greatest amount of water to flow. What is the maximum​ cross-sectional area?
Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0

Answer:

Depth of the rain gutter is 8 inches

Step-by-step explanation:

Let’s assume ‘x’ is the depth of the rain gutter

Then the width of the rain gutter can be written as 16 - 2x

Cross sectional area

A = depth x width

Substitute values

A = x*(16 - 2x)

A = 16x – 2x^2

Now according to axis of symmetry for maximum area x = -b/2a

x = -16/2*(-2)

x = 4 inches depth of rain gutter, substitute the value of x to get

Width of rain gutter 16 – 2(4) = 8 inches

Area of the rain gutter for maximum water flow

A = 4 * 8

A = 32 square inch.

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A direct variation function contains the points (2, 14) and (4, 28). Which equation represents the function?
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y=7x

Step-by-step explanation:

we know that

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we have the points (2,14) and (4,28)

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For (2,14) ----> k=14/2=7

For (4,28) ----> k=28/4=7

so                              

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Please help me ASAP
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Answer:

<em>The Graph is shown below</em>

Step-by-step explanation:

<u>The Graph of a Function</u>

Given the function:

\displaystyle y=g(x)=-\frac{3}{2}(x-2)^2

It's required to plot the graph of g(x). Let's give x some values:

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Point (0,-6)

\displaystyle y=g(2)=-\frac{3}{2}(2-2)^2=-\frac{3}{2}(0)^2=0

Point (2,0)

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Point (4,-6)

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Point (6,-24)

The graph is shown in the image below

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