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Ghella [55]
3 years ago
5

You can work no more than 60 hours each week at your two jobs. Dog walking pays $7 per hour and your sales job at Computers &amp

; More, Inc. pays $12 per hour. You need to earn at least $450 each week to pay your bills. Your friend solves the system of inequalities and tells you that a possible solution is (-3, 50). Is this a possible solution, why or why not?
Mathematics
2 answers:
DedPeter [7]3 years ago
5 0

Answer:

No

Step-by-step explanation:

You cannot work negative 3 hours, it's impossible. One possible solution would be to work 50 hours a week at dog walking for 7 dollars an hour, and 10 hours a week at Computers & More, Inc. for 12 dollars a week. This would give you 350 dollars from dog walking, and 120 dollars from Computers & More, Inc. This would be a total of 470 dollars.

jolli1 [7]3 years ago
4 0

Answer:

No, this is not a possible solution because it does not satisfy all inequalities of the system.

Step-by-step explanation:

Let x represents the number of hours spent on dog walking and y represents the number of hours spent on sales job at Computers & More Inc.,

Given,

Total hours can not be no more than 60 hours,

⇒ x + y ≤ 60

Dog walking pays $7 per hour and your sales job at Computers & More, Inc. pays $12 per hour.

Thus, the total earning = 7x + 12y

According to the question,

Total earning ≥ $ 450

⇒ 7x + 12y ≥ 450

Also, hours can not be negative,

⇒ x ≥ 0; y ≥ 0

Hence, the system of inequalities that shows the given situation is,

7x + 12y ≥ 450;  

x + y ≤ 60;

x ≥ 0; y ≥ 0

Since, -3 ≥ 0 ( False ),

Thus, the point is not satisfying all inequalities of the system,

Hence, it can not be the solution.

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Answer:  \bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}

<u>Step-by-step explanation:</u>

(1)           (12, 18, 27, ...)

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r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\  r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}

(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad  r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\  r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}

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Step-by-step explanation:

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