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Elan Coil [88]
3 years ago
9

Dr. George is ready to by a new iphone and has two different carriers he can go with. Cell 4 U can be represented by the functio

n c(x)=10x+75. Data House can be represented by the function c(x)=15x+50. X represents the amount of data used. Accessories will cost $135 Dr. George usually uses 4 gigs of data a month.
1. How much will the phone cost at Cell 4U?

2. How much will the phone cost at Data House?

3. Which company should Dr. George go with? Justify your answer.
Mathematics
1 answer:
Aleks04 [339]3 years ago
8 0
At Cell 4U it would cost $250 with accessories 
At Data House it would cost $245 with accessories.
He should buy a phone from Data house because it will be 5 dollars cheaper
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Answer:

number of units for cost to be minimum=150

Step-by-step explanation:

y=2x^2-600 x+49000

dy/dx=4x-600

dy/dx=0 gives 4x-600=0

4x=600

x=150

d^2y/dx^2=4x

at x=150,d^2y/dx^2=4*150=600>0

so y is minimum at x=150

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Step-by-step explanation:

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
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Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0
3 years ago
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