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3 years ago

I need help with this question to

Mathematics

2 answers:

madam [21]3 years ago

7 0

Well if you have a question, why dont you post the question....

Semenov [28]3 years ago

6 0

I believe that it would be A.Associative property

Please mark as brainliest answer

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A rectangular tank that is 5324 ft cubed with a square base and open top is to be constructed of sheet steel of a given thicknes

marishachu [46]

Answer:

Side of 22 and height of 11

Step-by-step explanation:

Let s be the side of the square base and h be the height of the tank. Since the tank volume is restricted to 5324 ft cubed we have the following equation:

$V = s^2h = 5324$

$h = 5324 / s^2$

As the thickness is already defined, we can minimize the weight by minimizing the surface area of the tank

Base area with open top $s^2$

Side area 4sh

Total surface area $A = s^2 + 4sh$

We can substitute $h = 5324 / s^2$

$A = s^2 + 4s\frac{5324}{s^2}$

$A = s^2 + 21296/s$

To find the minimum of this function, we can take the first derivative, and set it to 0

$A' = 2s - 21296/s^2 = 0$

$2s = 21296/s^2$

$s^3 = 10648$

$s = \sqrt[3]{10648} = 22$

$h = 5324 / s^2 = 5324 / 22^2 = 11$

4 0

3 years ago

PLEASE HELP Meeeeeeeeeeeeeeeeee

klemol [59]

The answer is A.Use a compound interest calculator. Multiple sites to use. For me, I had to use my head for a while

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3 years ago

Plz help: What is 275 time 9000.

Kamila [148]

2,475,000 this your answer

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2 years ago

Read 2 more answers

Describe the general properties of rotations.

Korolek [52]

A rotation maps a line to a line, a ray to a ray, a segment to a segment, and an angle to an angle.

A rotation preserves lengths of segments.

A rotation preserves measures of angles. Step-by-step explanation:

You Welcome :)!

5 0

3 years ago

A rectangular storage container with a lid is to have a volume of 2 m3. The length of its base is twice the width. Material for

Scilla [17]

Answer:

Dimensions are 2 m by 1 meter by 1 meter,

Minimum cost is $ 18.

Step-by-step explanation:

Let w be the width ( in meters ) of the container,

Since, the length is twice of the width,

So, length of the container = 2w,

Now, if h be the height of the container,

Volume = length × width × height

2 = 2w × w × h

1 = w² × h

$\implies h=\frac{1}{w^2}$

Since, the area of the base = l × w = 2w × w = 2w²,

Area of the lid = l × w = 2w²,

While the area of the sides = 2hw + 2hl

= 2h( w + l)

$= 2\times \frac{1}{w^2}(w+2w)$

$=\frac{6w}{w^2}$

$=\frac{6}{w}$

Since, Material for the base costs $1 per m². Material for the sides and lid costs $2 per m²,

So, the total cost,

$C(w) = 1\times 2w^2+2\times 2w^2 + 2\times \frac{6}{w}$

$=2w^2+4w^2+\frac{12}{w}$

$=6w^2+\frac{12}{w}$

Differentiating with respect to w,

$C'(w) = 12w -\frac{12}{w^2}$

Again differentiating with respect to w,

$C''(w) = 12 + \frac{24}{w^3}$

For maxima or minima,

C'(w) = 0

$\implies 12w -\frac{12}{w^2}=0$

$\implies 12w^3 - 12=0$

$w^3-1=0\implies w = 1$

For w = 1, C''(w) = positive,

Hence, for width 1 m the cost is minimum,

Therefore, the minimum cost is C(1) = 6(1)²+12 = $ 18,

And, the dimension for which the cost is minimum is,

2 m by 1 meter by 1 meter.

7 0

3 years ago