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Vinvika [58]
3 years ago
15

Which of the following situations can be represented by a function with an infinite number of values in its range and only a fin

ite number of values in its domain? A. The cost of a cell phone bill depends on the number of minutes spent talking on the phone. For one billing period, the customer has a maximum of 100 minutes to use. B. The cost of a cell phone bill depends on the number of minutes spent talking on the phone. Regardless of the number of minutes, the bill can be at most $100. C. The cost of a cell phone bill depends on the number of minutes spent talking on the phone. For every minute used on the phone, the cost is $.15. D. It is not possible for a function to have an infinite number of values in its range and only a finite number of values in its domain.
Mathematics
1 answer:
PIT_PIT [208]3 years ago
7 0
B...............................
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Answer:

C

Step-by-step explanation:

The scale factor is the ratio of corresponding sides, image to original, so

scale factor = \frac{EF}{BC} = \frac{11}{66} = \frac{1}{6} → C

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How do you divide 15 pounds of rice into 4 unequal measuring using ounces
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F(x)=−2x+16<br> g(x)=4x+10<br> please give me the intersection points
vagabundo [1.1K]

Answer:

  • (1, 14)

Step-by-step explanation:

The intersection is when both functions have same coordinates

  • f(x) = g(x)

Substitute to get

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  • 4x + 2x = 16 - 10
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The y- coordinate is

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So the intersection point is (1, 14)

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2 years ago
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Money Flow  The rate of a continuous money flow starts at $1000 and increases exponentially at 5% per year for 4 years. Find the
77julia77 [94]

Answer:

Present value =  $4,122.4

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Step-by-step explanation:

Data provided in the question:

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Time = 4 years

Interest rate = 3.5%  compounded continuously

Now,

Accumulated Value of the money flow = 1000e^{0.05t}

The present value of the money flow = \int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt

= 1000\int\limits^4_0 {e^{0.015t}} \, dt

= 1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4

= 1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right]

= 1000 × [70.7891 - 66.6667]

= $4,122.4

Accumulated interest = e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt

= e^{0.035\times4}\times4,122.4

= $4,742

8 0
3 years ago
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Snezhnost [94]

Answer:

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Step-by-step explanation:

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3 years ago
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