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IgorC [24]
3 years ago
9

Jeff subscribed to cable for 75.00 a month and 140.00 installation andrew has another cable co. For 20.00 installation and 90.00

a month. In how many months will we be paying the same amount for their service.? And what will be the amount they pay?
Mathematics
1 answer:
lilavasa [31]3 years ago
7 0
8 months; $740

75x+140=90x+20
Find x then plug it in to find the cost
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Divide common factors from both numerator and denominator.

Divide 2

8/2 = 4

10/2 = 5

8/10 simplified = 4/5

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Answer:

The answer is D: 2+2+b=12

Step-by-step explanation:

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Which geometric figure are always paraelograms
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2 years ago
A car can be rented for $75 per week plus $0.35 per mile. How many miles can be driven if you have at most $110 to spend for wee
sergey [27]

After you subtract the fixed cost of $75 from your budget of $110, you have $35 to spend on miles. At $0.35 per mile, you can afford 100 of them.

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3 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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