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Brums [2.3K]
3 years ago
10

A young woman wants to make at least $200 a week and can't work no more than 30 hours a week. she works at the library for $8 an

hour and babysits for $6 an hour.
a. What system of inequalities shows the possible combination of hours and jobs she can work?
b. Why did you exclude points to the left of the y-axis and below the x-axis
Mathematics
1 answer:
Bas_tet [7]3 years ago
5 0
Let, Number of Hours in Library = x
Number of Hours in babysitting = y

Inequality would be: x + y = 30  ;
 &   8x + 6y ≥ 200

b) Solving: Multiply first equation by 6, 
6x + 6y ≥ 180
Subtract it from 2nd equation, 
2x ≥ 20
x ≥ 10

Now, substitute it in 1st equation, 
10 + y = 30
y = 20

Your Answer would be (10, 20)
x - value is lesser or equal to 10, so it would move to left (decreasing direction)!

Hope this helps!
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What you know:

Mr. Adams spent $36.25 on 12.5 pounds.

All you do to find the cost per each pound is:

36.25/12.5

Your answer: 
$2.9 dollars per pound.
8 0
3 years ago
Wright the expression 6a - 2 (a-1) in simplest form
vovikov84 [41]
6a-2(a-1)
6a-2a-2
4a-2
I think it is 4a-2
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3 years ago
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7 0
3 years ago
6. Give examples of methods to safeguard inventory.
snow_lady [41]

Answer:

1. Security Technology. ...

2. Security Personnel. ...

3. Inventory Audits. ...

4. Just-in-Time Inventory.

Step-by-step explanation:

Reviewing a few examples of safeguarding inventory can shed light on common inventory security methodologies, helping you to implement the ideal inventory safeguards for your business.

Security Technology. ...

Security Personnel. ...

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5 0
2 years ago
A King in ancient times agreed to reward the inventor of chess with one grain of wheat on the first of the 64 squares of a chess
statuscvo [17]
Let's start by visualising this concept.

Number of grains on square:
1   2   4   8   16 ...

We can see that it starts to form a geometric sequence, with the common ratio being 2.

For the first question, we simply want the fifteenth term, so we just use the nth term geometric form:
T_n = ar^{n - 1}
T_{15} = 2^{14} = 16384

Thus, there are 16, 384 grains on the fifteenth square.

The second question begs the same process, only this time, it's a summation. Using our sum to n terms of geometric sequence, we get:
S_n = \frac{a(r^{n} - 1)}{r - 1}
S_{15} = \frac{2^{15} - 1}{2 - 1}
S_{15} = 2^{15} - 1 = 32767

Thus, there are 32, 767 total grains on the first 15 squares, and you should be able to work the rest from here.
6 0
3 years ago
Read 2 more answers
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