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Bumek [7]
4 years ago
13

A cable company charges $75 for installation plus $20 per month. Another cable company offers free installation but charges $35

per month. For how many months of cable service would the total cost from either company be the same?
(a) 5
(b) 7
(c) 2
(d) 3
Mathematics
2 answers:
uranmaximum [27]4 years ago
7 0
Your answer would be (A) :)
kifflom [539]4 years ago
3 0
It would be 5 months because if you do 75 + 20x = 35x
15x = 75
----- ------
15 15
x = 5
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Answer:

Step-by-step explanation:

hello :

√(5/4)= √5/ √4 =√5/2 = (1/2)√5

3+√5/4−√5 = 3+(1/2)√5   so : a=3  and  b= 1/2

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Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shap
pantera1 [17]

Answer:

The height of the pile is increasing at the rate of  \mathbf{ \dfrac{20}{56.25 \pi}   \ \ \ \ \  ft/min}

Step-by-step explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e \dfrac{dV}{dt}= 20 \ ft^3/min

we know that radius r is always twice the   diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

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V = \dfrac{\pi r^2 h}{3}

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Taking the differentiation of volume V with respect to time t; we have:

\dfrac{dV}{dt }= (\dfrac{d}{dh}(\dfrac{\pi h^3}{12})) \times \dfrac{dh}{dt}

\dfrac{dV}{dt }= (\dfrac{\pi h^2}{4} ) \times \dfrac{dh}{dt}

we know that:

\dfrac{dV}{dt}= 20 \ ft^3/min

So;we have:

20= (\dfrac{\pi (15)^2}{4} ) \times \dfrac{dh}{dt}

20= 56.25 \pi \times \dfrac{dh}{dt}

\mathbf{\dfrac{dh}{dt}= \dfrac{20}{56.25 \pi}   \ \ \ \ \  ft/min}

The height of the pile is increasing at the rate of  \mathbf{ \dfrac{20}{56.25 \pi}   \ \ \ \ \  ft/min}

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bija089 [108]

Answer:

-3x^{4} + 19x^{3} - 38x^{2} + 25x - 3

Step-by-step explanation:

1) distribute x² into (-3x² + 7x - 1) to get: -3x^{4} + 7x³ - x²

2) distribute -4x into (-3x² + 7x - 1) to get: 12x³ - 28x + 4x

3) distribute 3 into (-3x² + 7x - 1) to get: -9x² + 21x - 3

4) combine all the answers together into one equation:

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5) combine like terms:

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6) combine answers together into one equation:

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I hope this helps you

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