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Paladinen [302]
3 years ago
11

Brainliest + Points Please explain..

Mathematics
1 answer:
prohojiy [21]3 years ago
8 0

Answer:

0\leq x\leq42

Step-by-step explanation:

We have been given that checking account earnings at Long's Bank are expressed by the equation I=-0.06x+8.3. The earnings at Fellow's Bank are modeled by I=-0.02x+6.6. In both cases x is the number of checks written.

To find the number of checks that will generate more earnings at Long's bank then Fellow's Bank, we will set an inequality as:

-0.06x+8.3>-0.02x+6.6    

Now let us solve for x.

-0.06x+0.02x+8.3>-0.02x+0.02x+6.6

-0.04x+8.3>6.6

-0.04x+8.3-8.3>6.6-8.3

-0.04x>-1.7    

Now let us divide both sides of our inequality by -0.04. Since dividing by a negative number swaps sign of inequality, so our inequality will be:

\frac{-0.04x}{-0.04}

x

Since the number of checks needs to be integer, so number of checks will be:

0\leq x\leq42

Therefore, the number of checks between 0 to 42 checks will generate more earnings income at Long's Bank than Fellow's Bank.

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Answer:

D) 1/27

Step-by-step explanation:

The question is: 3^(-3).

Note that there is a negative sign in the number at the power place. This means that you must flip the number, and in this case, put it over 1. The rest is solved as usual:

3^-3 = 1/(3^3) = 1/(3 * 3 * 3) = 1/27

1/27, or D) is your answer.

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Find the inverse laplace transform of: (2 s + 4) / (s - 3)^3
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Answer:

e^{3t}(2t+5t^{2})

Step-by-step explanation:

L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=

Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3

Translation theorem:L^{1} [F(s-a)=L^{-1}[F(s)|_{s \to s-a}\\ L^{1} [F(s-a)=e^{at} f(t)

L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=e^{3t} L^{-1}[\frac{2(s+3)+4}{s^{3}} ]

Separate the fraction in a sum:

e^{3t} L^{-1}[\frac{2s+10}{s^{3}} ]=e^{3t} L^{-1}[\frac{2s}{s^{3}}+\frac{10}{s^{3}} ]=e^{3t} (L^{-1}[\frac{2}{s^{2}}]+ L^{-1}[\frac{10}{s^{3}}])

The formula for this is:

L^{-1}[\frac{n!}{s^{n+1}} ]=t^{n}

Modify the expression to match the formula.

e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ \frac{10}{2} L^{-1}[\frac{2}{s^{2+1}}])=e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])

Solve

e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])=e^{3t}(2t+5t^{2} )

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