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Stells [14]
3 years ago
11

You have a part-time job. One day your boss offers to pay you either 2h−1 or 2h − 1 dollars for each hour h you work that day. C

omplete the table. Which option should you choose? Explain.
I filled in the first part of the table I just need help on the second part.

Mathematics
2 answers:
liberstina [14]3 years ago
4 0
For the second part, you have to subtract 1 from the number of hours and raise it to that power.

For example, when h = 1, you have:

2^(1 - 1)
2^0
1

The 5 values in your table would be:
1  2  4  8  16


AlekseyPX3 years ago
4 0

Answer:

The better option is 2^h -1 hour every h hour, because in this way, we can earn more.  

Step-by-step explanation:

We are given the following information in the question:

On a part time job, a person is paid 2^{h-1} for every hour h he works.

Then payment can be written as:

h = 1, 2^{h-1} = 2^{1-1} = 2^0 = 1\\h = 2, 2^{h-1} = 2^{2-1} = 2^1 = 2\\h = 3, 2^{h-1} = 2^{3-1} = 2^2 = 4\\h = 4, 2^{h-1} = 2^{4-1} = 2^3 = 8\\h = 5, 2^{h-1} = 2^{5-1} = 2^4 = 16

Table:

   h:       1        2        3        4         5

Pay:      1         2        4       8          16

The better option is 2^h -1 hour every h hour, because in this way, we can earn more.

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I need to solve the equation please help me
lesantik [10]

Answer:

\displaystyle x=\Big\{\frac{7\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi, \frac{11\pi}{6}+2n\pi}\Big\}, n\in\mathbb{Z}

Step-by-step explanation:

We are given the equation:

4\sin^2(x)+6\sin(x)+2=0

First, we can divide everything by 2:

2\sin^2(x)+3\sin(x)+1=0

Notice that we have an equation in quadratic form. Namely, if we make a substitution where u = sin(x), we acquire:

2u^2+3u+1=0

Solve for u. Factor:

(2u+1)(u+1)=0

Zero Product Property:

2u+1=0\text{ or } u+1=0

Solving for both cases:

\displaystyle u=-\frac{1}{2}\text{ or } u=-1

And by substitution:

\displaystyle \sin(x)=-\frac{1}{2}\text{ or } \sin(x)=-1

For the first case, recall that sin(x) is -1/2 for every 7π/6 and every 11π/6. Hence, for the first case, our solutions are:

\displaystyle x=\frac{7\pi}{6}+2n\pi \text{ and } x=\frac{11\pi}{6}+2n\pi, n\in\mathbb{Z}

Where n is an integer.

For the second case, sin(x) is -1 for every 3π/2. Thus:

\displaystyle x=\frac{3\pi}{2}+2n\pi

All together, our solutions are:

\displaystyle x=\Big\{\frac{7\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi, \frac{11\pi}{6}+2n\pi}\Big\}, n\in\mathbb{Z}

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