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3 years ago

Transparency and worksheet master lesson 4.2 page 166

Mathematics

1 answer:

arlik [135]3 years ago

3 0

Http://mrdealy.weebly.com/uploads/2/3/5/2/23520438/geom._ch._3_angle_diagrams__3.8_worksheet.pdf

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The length of a rectangle is 24 units. Can the perimeter x of the rectangle be 60 units when its width y is 11 units? A. No, the

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Answer:

Step-by-step explanation:

The formula for the Perimeter of a rectangle is P = 2W + 2L.

Let's equate this to 60 units and substitute 11 for W and 24 units for L:

2(11 units) + 2(24 units) = 22 units + 48 units = 70 units. NO (Answer A)

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3 years ago

Factor completely x8 − 625.

Zina [86]

The answer is D) (x2 − 5)(x2 + 5)(x4 + 25)

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A box with a square base and no top is to be built with a volume of 1638416384 in33. Find the dimensions of the box that require

julsineya [31]

Answer:

$512\ \text{in}^2$

Step-by-step explanation:

x = Length and width of base

y = Height of box

Volume of the box is $16384\ \text{in}^3$

$x^2y=16384\\\Rightarrow y=\dfrac{16384}{x^2}$

Surface area is given by

$s=x^2+4y\\\Rightarrow s=x^2+4\times \dfrac{16384}{x^2}\\\Rightarrow s=x^2+\dfrac{65536}{x^2}$

Differentiating with respect to x we get

$s'=2x-\dfrac{131072}{x^3}$

Equating with 0 we get

$0=2x^4-131072\\\Rightarrow x=(\dfrac{131072}{2})^{\dfrac{1}{4}}\\\Rightarrow x=16$

$s''=2+\dfrac{393216}{x^4}$

at $x=16$

$s''=2+\dfrac{393216}{16^4}=8>0$

So the function is minimum at x = 16

$y=\dfrac{16384}{x^2}=\dfrac{16384}{16^2}\\\Rightarrow y=64$

The material required is

$s=x^2+4y=16^2+4\times 64\\\Rightarrow s=512\ \text{in}^2$

The minimum amount of material required is $512\ \text{in}^2$ .

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3 years ago

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Answer:

d = |145 − (−235)|

Step-by-step explanation:

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Oder these number from least to greatest -3, 5,-1, 0, 7

kotykmax [81]

Answer:

-3,-1,0,5,7

Step-by-step explanation:

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