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2 hours ago

6

The ideal width of a certain conveyor belt for a manufacturing plant is 50 in. Conveyor belts can vary from the ideal width by a

t most 7/32in. Which of the following represents the acceptable widths for the conveyor belt? Check all that apply.
x>=49 25/32
x<=50 7/32
49 25/32<=x<=50 7/32
x+7/32<=50
|x-50|< 7/32

1 answer:

Zina [3.8K]2 hours ago

7 0

Answer:

The answer is:

$|x-50|\leq \dfrac{7}{32}$

$49\dfrac{25}{32}\leq x\leq 50\dfrac{7}{32}$

Step-by-step explanation:

The preferred width for a conveyor belt used in a manufacturing facility is 50 inches.

These conveyor belts can differ from this optimal width by no more than 7/32 inch.

This means the change in width must be equal to or less than 7/32 inch and cannot go beyond that limit.

To illustrate, if it decreases, the permitted width is shown with the expression;

$|x-50|\leq \dfrac{7}{32}$

Solving this gives us:

$\dfrac{-7}{32}\leq x-50\leq \dfrac{7}{32}\\\\\\50-\dfrac{7}{2}\leq x\leq 50+\dfrac{7}{2}\\\\49\dfrac{25}{32}\leq x\leq 50\dfrac{7}{32}$

Thus, the correct selections are:

$|x-50|\leq \dfrac{7}{32}$

$49\dfrac{25}{32}\leq x\leq 50\dfrac{7}{32}$

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Response-

(78,104) represents the point closest to the interior.

Explanation-

The equation defining the circle,

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Since the point lies on the circle, its coordinates must be,

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The distance "d" from the point to (30,40) can be calculated as,

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$\Rightarrow 80x=60\sqrt{16900-x^2}$

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$\Rightarrow 6400x^2=3600(16900-x^2)$

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Since f''(x) is positive, the function f(x) achieves its minimum at x=78

When x is set to 78, the corresponding y value will be

$\Rightarrow y = \sqrt{16900-x^2}=\sqrt{16900-78^2}=104$

This leads us to conclude that the closest point is (78,104)

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