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3 years ago

Determine the perimeter of ABCD.

Mathematics

1 answer:

Galina-37 [17]3 years ago

7 0

You add of the sides to get the perimeter.. 9 plus 9 is 18 and 5 plus 5 is 10. 10 plus 18 of 28. The perimeter of ABCD is 28.
I hope this helps.

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A square has an area that is less than 100 m. What is a reasonable range for the graph of the

WITCHER [35]

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3 years ago

16^3x+8^x+6 Could you please give me the steps as well for this problem?

BARSIC [14]

Answer:

x = 2

Step-by-step explanation:

$16^{3x} = 8^{x+6} \\2^{4(3x)} = 2^{3(x+6)} -->$ (16 = $2^{4}$ and 8 = $2^{3}$ )

$2^{12x} = 2^{3x+18}$

Since the base number is 2, the exponents must equal each other.

So:

12x = 3x+18

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2 years ago

Jasmine sells bracelets at a local jewelry store in South Dekalb Mall. She has 1,190 purple beads and 2,125 pink beads. Each bra

jeka57 [31]

Answer:

Step-by-step explanation:

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2 years ago

(3 pt) point v is graphed on a number line at –6. point u is 8 units away from point v on the number line. what could be the coo

lukranit [14]

Point v is graphed on a number line at -6...point u is 8 units away from v....notice that it doesn't say which way....so it can go either way....so we need to find 8 units away from -6 in both directions

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3 0

3 years 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