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3 years ago

Can i get help please ?

Mathematics

1 answer:

wariber [46]3 years ago

4 0

Answer:

D -- the last one

Step-by-step explanation:

The formula tells you what to do.

t = 3

A0 = 25

A(t) = ?

A(3) = 25 * (1/2)^3

A(t) = 25 * 1/8

A(t) = 25/8

A(t) = 3.125

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2 years ago

If y is 2.5 when x is 5 and y varies directly with x, find y when x is 14. 7 7.5 12.5 28

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3 years ago

Consider the function V=g(x), where g(x) =x(6-2x)(8-2x), with x being the length of a cutout in cm and V being the volume of an

Andrej [43]

Answer:

The maximum volume of the open box is 24.26 cm³

Step-by-step explanation:

The volume of the box is given as $V=g(x)$ , where $g(x)=x(6-2x)(8-2x)$ and $0\le x\le3$ .

Expand the function to obtain:

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$g'(x)=12x^2-56x+48$

To find the point where the maximum value occurs, we solve

$g'(x)=0$

$\implies 12x^2-56x+48=0$

$\implies x=1.13,x=3.54$

Discard x=3.54 because it is not within the given domain.

Apply the second derivative test to confirm the maximum critical point.

$g''(x)=24x-56$ , $g''(1.13)=24(1.13)-56=-28.88\:$

This means the maximum volume occurs at $x=1.13$ .

Substitute $x=1.13$ into $g(x)=x(6-2x)(8-2x)$ to get the maximum volume.

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The maximum volume of the open box is 24.26 cm³

See attachment for graph.

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3 years ago

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wariber [46]

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3 years ago

Read 2 more answers

HELP PLEASE 16 points

Lina20 [59]

Answer: 21

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3 years ago