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2 years ago

Find the perimeter of the following figure

Mathematics

1 answer:

Nastasia [14]2 years ago

4 0

I believe it’s 32.7 ksmakksmakaomwkaimwnsiaklalaowkwnne

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Which number has the least value?

Vlad1618 [11]

Answer:

Your annswer is b

Step-by-step explanation: because it is least than all of the others

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2 years ago

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Help me plz it’s hard for me lol

slava [35]

Answer:

Step-by-step explanation:

2 *3 sqare root 3 * 12

Solving gives:

6 sqare root 36.

sqare root 36 = 6

2 * 6 * 3

So our answer is 36.

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3 years ago

Which side lengths form a right triangle?

kakasveta [241]

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Step-by-step explanation:

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2 years ago

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Which graph represents the solution 8 - 3x ≥ -4

Vedmedyk [2.9K]

Answer:

3rd option

Step-by-step explanation:

8 - 3x ≥ - 4 ( subtract 8 from both sides )

- 3x ≥ - 12

divide both sides by - 3 , reversing the symbol as a result of dividing by a negative quantity.

x ≤ 4

since ≤ the circle on number line is solid ( not open )

since less than 4 then arrow points left away from 4

the solution is shown on the third graph

6 0

1 year ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the bin is constructed from 48 square feet of sheet metal, so:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

2 years ago

Find the perimeter of the following figure​

Find the perimeter of the following figure