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jekas [21]
3 years ago
9

The function f(t) represents the cost to connect to the Internet at an online gaming store. It is a function of t, the time in m

inutes spent on the Internet.
f(t) =StartLayout Enlarged left-brace 1st row 1st column 0 dollars 2nd column 0 less-than t less-than-or-equal-to 30 2nd row 1st column 5 dollars 2nd column 30 less-than t less-than-or-equal-to 90 3rd row 1st column 10 dollars 2nd column t greater-than 90 EndLayout

Which statement is true about the Internet connection cost?

It costs $5 per hour to connect to the Internet at the gaming store.
The first half hour is free, and then it costs $5 per minute to connect to the Internet.
It costs $10 for each 90 minutes spent connected to the Internet at the gaming store.
Any amount of time over an hour and a half would cost $10.
Mathematics
2 answers:
dolphi86 [110]3 years ago
3 0

Answer:

<u>The correct answer is D. Any amount of time over an hour and a half would cost $10.</u>

Step-by-step explanation:

f (t), when t is a value between 0 and 30

The cost is US$ 0 for the first 30 minutes

f (t), when t is a value between 30 and 90

The cost is US$ 5 if the connection takes between 30 and 90 minutes

f (t), when t is a value greater than 90

The cost is US$ 10 if the connection takes more than 90 minutes

According to these costs, statements A, B and C are incorrect. The connection doesn't cost US$ 5 per hour like statement A affirms, the cost of the connection isn't US$ 5 per minute after the first 30 minutes free as statement B affirms and neither it costs US$ 10 for every 90 minutes of connection, as statement C affirms. <u>The only one that is correct is D, because any amount of time greater than 90 minutes actually costs US$ 10.</u>

irinina [24]3 years ago
3 0

Answer:

The correct answer is D. Any amount of time over an hour and a half would cost $10.

Step-by-step explanation:

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Let f(x) = -3x + 2 and g(x) = 2x – 5;<br> compute g(f(x))
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Step-by-step explanation:

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f(x) = -3x + 2 and g(x) = 2x – 5

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g(f(x)) = 2(-3x + 2)+2

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Using Laplace transforms, solve x" + 4x' + 6x = 1- e^t with the following initial conditions: x(0) = x'(0) = 1.
professor190 [17]

Answer:

The solution to the differential equation is

X(s)=\cfrac 1{6}  -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)

Step-by-step explanation:

Applying Laplace Transform will help us solve differential equations in Algebraic ways to find particular  solutions, thus after applying Laplace transform and evaluating at the initial conditions we need to solve and apply Inverse Laplace transform to find the final answer.

Applying Laplace Transform

We can start applying Laplace at the given ODE

x''(t)+4x'(t)+6x(t)=1-e^t

So we will get

s^2 X(s)-sx(0)-x'(0)+4(sX(s)-x(0))+6X(s)=\cfrac 1s -\cfrac1{s-1}

Applying initial conditions and solving for X(s).

If we apply the initial conditions we get

s^2 X(s)-s-1+4(sX(s)-1)+6X(s)=\cfrac 1s -\cfrac1{s-1}

Simplifying

s^2 X(s)-s-1+4sX(s)-4+6X(s)=\cfrac 1s -\cfrac1{s-1}

s^2 X(s)-s-5+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}

Moving all terms that do not have X(s) to the other side

s^2 X(s)+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}+s+5

Factoring X(s) and moving the rest to the other side.

X(s)(s^2 +4s+6)=\cfrac 1s -\cfrac1{s-1}+s+5

X(s)=\cfrac 1{s(s^2 +4s+6)} -\cfrac1{(s-1)(s^2 +4s+6)}+\cfrac {s+5}{s^2 +4s+6}

Partial fraction decomposition method.

In order to apply Inverse Laplace Transform, we need to separate the fractions into the simplest form, so we can apply partial fraction decomposition to the first 2 fractions. For the first one we have

\cfrac 1{s(s^2 +4s+6)}=\cfrac As + \cfrac {Bs+C}{s^2+4s+6}

So if we multiply both sides by the entire denominator we get

1=A(s^2+4s+6) +  (Bs+C)s

At this point we can find the value of A fast if we plug s = 0, so we get

1=A(6)+0

So the value of A is

A = \cfrac 16

We can replace that on the previous equation and multiply all terms by 6

1=\cfrac 16(s^2+4s+6) +  (Bs+C)s

6=s^2+4s+6 +  6Bs^2+6Cs

We can simplify a bit

-s^2-4s=  6Bs^2+6Cs

And by comparing coefficients we can tell the values of B and C

-1= 6B\\B=-1/6\\-4=6C\\C=-4/6

So the separated fraction will be

\cfrac 1{s(s^2 +4s+6)}=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6}

We can repeat the process for the second fraction.

\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac A{s-1} + \cfrac {Bs+C}{s^2+4s+6}

Multiplying by the entire denominator give us

1=A(s^2+4s+6) + (Bs+C)(s-1)

We can plug the value of s = 1 to find A fast.

1=A(11) + 0

So we get

A = \cfrac1{11}

We can replace that on the previous equation and multiply all terms by 11

1=\cfrac 1{11}(s^2+4s+6) + (Bs+C)(s-1)

11=s^2+4s+6 + 11Bs^2+11Cs-11Bs-11C

Simplifying

-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C

And by comparing coefficients we can tell the values of B and C.

-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C\\-1=11B\\B=-\cfrac{1}{11}\\5=-11C\\C=-\cfrac{5}{11}

So the separated fraction will be

\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac {1/11}{s-1} + \cfrac {-s/11-5/11}{s^2+4s+6}

So far replacing both expanded fractions on the solution

X(s)=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6} -\cfrac {1/11}{s-1} -\cfrac {-s/11-5/11}{s^2+4s+6}+\cfrac {s+5}{s^2 +4s+6}

We can combine the fractions with the same denominator

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {-s/6-4/6+s/11+5/11+s+5}{s^2 +4s+6}

Simplifying give us

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{s^2 +4s+6}

Completing the square

One last step before applying the Inverse Laplace transform is to factor the denominators using completing the square procedure for this case, so we will have

s^2+4s+6 = s^2 +4s+4-4+6

We are adding half of the middle term but squared, so the first 3 terms become the perfect  square, that is

=(s+2)^2+2

So we get

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{(s+2)^2 +(\sqrt 2)^2}

Notice that the denominator has (s+2) inside a square we need to match that on the numerator so we can add and subtract 2

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61(s+2-2)/66+316 /66}{(s+2)^2 +(\sqrt 2)^2}\\X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66+194 /66}{(s+2)^2 +(\sqrt 2)^2}

Lastly we can split the fraction one more

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66}{(s+2)^2 +(\sqrt 2)^2}+\cfrac {194 /66}{(s+2)^2 +(\sqrt 2)^2}

Applying Inverse Laplace Transform.

Since all terms are ready we can apply Inverse Laplace transform directly to each term and we will get

\boxed{X(s)=\cfrac 1{6}  -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)}

6 0
3 years ago
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