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3 years ago

NEED HELP ASAP! Plz answer all the questions! Including the one on the picture! Will give brainlest!

Mathematics

2 answers:

mote1985 [20]3 years ago

8 0

Answer:

A) 2/3 or 0.6 repeating

B) 17

C) 5

D) -17/4 or -4.25 or -4 1/4

E) 3

xeze [42]3 years ago

4 0

-2/3-(-1 1/3) =

-2/3 + 1 1/3 =

4/3 - 2/3 =

2/3

12 - (-5) =

12 + 5 =

17

-1 - (-6) =

-1 + 6 =

6 - 1 =

5

-3 3/8 - 7/8 =

27/8 - 7/8 =

20/8 =

2 4/8 =

2 1/2

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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:

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

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3 years ago

Integrate with respect to x 1/√(1-2x)

Ilia_Sergeevich [38]

Answer:

-√(1 - 2x) + C

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1/√(1-2x)

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-½∫1/√(u) du = -√(u) + C

Plugging in the value of u, we will have;

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3 years ago