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1 answer:

meriva2 years ago

6 0

Answer:

if im not wrong its 1 and 2

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in a right triangle the length of a hypotenuse is c and the length of one leg is a, and the length of the other leg is b what is

puteri [66]

Answer:

Step-by-step explanation:

a=12 (height)

c=13 (hypotenous)

b=? (base)

By using Pythagoras theorm

We have, hypotenous sq=base sq+height sq

Let b =base SO, base sq=hypotenous Sq-height Sq

=>base Sq =13sq-12sq

Base Sq =169-144

Base Sq =25

Base =5 (shifting the sq)....

Hope it's helpful

5 0

2 years ago

Can anyone answer this word problem for me?

mel-nik [20]

What’s the word problem

8 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 <span>bin is constructed from 48 square feet of sheet metal, s</span>o:

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, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

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