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Paha777 [63]
3 years ago
10

PLEASE HELP!!!!!!!

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
5 0

Answer:

C

Step-by-step explanation:

Viefleur [7K]3 years ago
5 0
A the answer is A because they have the same ending
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If i walk 2 block in 15 minutes what will my speed be?
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Answer:

Your answer would be 6mph.

Step-by-step explanation:

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Emery borrowed money from her brother to buy a new phone, and is paying off a fixed amount each week. After 2 weeks, she will ow
jarptica [38.1K]

Answer: A. The original price was $608. B. $76 each week C. Quite useful, as it keeps track of the amount of money she still needs to owe each week.

Step-by-step explanation: For part A, 5 week subtracted by 2 weeks is 3 weeks. 456-228 is 228. Then divide 228 by 3 and you get $76. Then do 2x76 to get 152 and add it to 456, and you get 608.

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Find the missing side lengths
timama [110]

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8 0
4 years ago
A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot
ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box  is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = (x\times x)

                               =x^2

The volume of the box  is = area of the base × height

                                           =x^2h

Therefore,

x^2h=272

\Rightarrow h=\frac{272}{x^2}

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

=20x^2 cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

=10x^2 cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

=4xh\times 1.5 cents

=6xh cents

Total cost = (20x^2+10x^2+6xh)

                =30x^2+6xh

Let

C=30x^2+6xh

Putting the value of h

C=30x^2+6x\times \frac{272}{x^2}

\Rightarrow C=30x^2+\frac{1632}{x}

Differentiating with respect to x

C'=60x-\frac{1632}{x^2}

Again differentiating with respect to x

C''=60+\frac{3264}{x^3}

Now set C'=0

60x-\frac{1632}{x^2}=0

\Rightarrow 60x=\frac{1632}{x^2}

\Rightarrow x^3=\frac{1632}{60}

\Rightarrow x\approx 3

Now C''|_{x=3}=60+\frac{3264}{3^3}>0

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is =\frac{272}{3^2}

                                          =30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0
3 years ago
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