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kirill115 [55]
3 years ago
13

Paul is comparing two high-speed Internet service providers (ISPs) for his business. The graph represents the cost of each ISP a

s a function of t, the duration in months of a service contract.
A coordinate grid showing Internet Service Provider Cost, with Service Contract in months along the horizontal axis, and Total Cost in dollars along the y-axis. Two lines are graphed. One line is labeled I S P A and passes through (0, 180) and (12, 300). The second line is labeled I S P B and passes through (0, 0) and (12, 300).

If C(t) = 180 + 10t represents ISP A and C(t) = 25t represents ISP B, how long would the service contracts need to be for the total costs to be the same?

6 months
8 months
10 months
12 months
Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
7 0

Answer:

12 months

Step-by-step explanation:

Theres an awful amount of needless information to get through.

Only the last paragrph is needed.

C(t) = 180 + 10t

C(t) = 25t

Set the equations equal

25t = 180 + 10t

Subtract 10t from both sides

15t = 180

Divide both sides by 15

t = 12

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Find the functional values g (-3), g (0), and g (5) for the compound function.
Zinaida [17]

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g(x) =\begin{cases}7 & \text{if } x \leq 0 \\ \\\dfrac{1}{x} & \text{if } x > 0\end{cases}

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Find the indefinite integral. (Use C for the constant of integration.) <br> e2x 25 e4x dx.
Sladkaya [172]

Answer:

The solution is  \frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] +  C

Step-by-step explanation:

From the question

    The function given is  f(x) =  \frac{e^{2x}}{ 25 + e^{4x}} dx

The  indefinite integral is  mathematically represented as

          \int\limits  {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx

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=>   \frac{du}{dx} 2e^{2x}

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=  \frac{1}{2} \int\limits  {\frac{1}{ 5^2 + u^2)} } \, du

= \frac{1}{2} \frac{tan^{-1} [\frac{u}{5} ]}{5}  +  C

Now substituting for  u

\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] +  C

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