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coldgirl [10]
2 years ago
13

Given the midpoint and one endpoint, find the other endpoint. Endpoint (2,-9) Midpoint (-2,-1)

Mathematics
1 answer:
Lynna [10]2 years ago
3 0

Step-by-step explanation:

Formula for midpoint is ;

m.p = ( x1 + x2/2 , y1 + y2/2)

So x1= 2 and y1 = -9 and the midpoint has already been given so just equate .

( -2 , -1 ) = ( 2 + x2/2 , -9 + y2 /2)

so now equate -2 to 2+x2/2 and -1 to -9 + y2/2

-2 = 2 + x2/ 2. -1 = -9 + y2 / 2

2 + x2 = -4. -9 + y2 = -2

x2 = -4 -2. y2 = -2 + 9

x2 = -6. y2 = 7

Answer:

( -6 , 7 )

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a box with a square base and open top must have a volume of 70304 c m 3 cm3 . we wish to find the dimensions of the box that min
Dmitriy789 [7]

The box should have base 52 cm by 52 cm and height 26 cm.

What is volume?

A three-dimensional space's occupied volume is measured. It is frequently expressed as a numerical value using SI-derived units, other imperial units, or US customary units. Volume definition and length definition are connected.

The Volume of a box with a square base x by x cm and height h cm is

V = x^2h

Since the surface area directly affects the amount of material used, we can reduce the amount of material by reducing the surface area.

The surface area of the box described is  A = x^2 + 4xh

We need A as a function of x alone, so we'll use the fact that

V = x^2h= 70304 cm^3

⇒ h = \frac{70304}{x^2}

So, the area becomes,

A = x^2+4x(\frac{70304}{x^2})\\ A = x^2 + \frac{281216}{x}

We want to minimize A, so

A' = 2x - \frac{281216}{x^2} =0\\\frac{2x^3-281216}{x^2}=0\\ {2x^3-281216} = 0\\x^3 - 140608=0\\x^3 = 140608\\x = 52

The second derivative test verifies that A has a minimum at this critical number:

A'' = 2+\frac{562432}{x^3}  which is positive at x = 52.

Now,

h = \frac{70304}{x^2}= \frac{70304}{(52)^2}=26

Hence, The box should have base 52 cm by 52 cm and height 26 cm.

To know more about volume, click on the link

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