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BartSMP [9]
1 year ago
14

Photo attached A. -2 B. -1 C. 1/-2 D. 1/2

Mathematics
1 answer:
kifflom [539]1 year ago
6 0

Step-by-step explanation:

(X1,Y1) = (-2,2)

(X2,Y2) = (0,1)

• Find slope.

m = (Y2 - Y1)/(X2 - X1)

m = (1 - 2)/(0 - (-2))

m = -1/2

The answer is C.

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Really need this answered :/ thank you
VladimirAG [237]

Answer:

i believe its the third one don't quote me on that though

Step-by-step explanation:


5 0
3 years ago
Change 7/5 to a whole number
Harlamova29_29 [7]
Write it as a mixed fraction 7/5 = 1 2/5
5 0
3 years ago
Calculate produce of 1/6 and 12/13
Sergeu [11.5K]

Answer:

1/6×12/13

2/13

Step-by-step explanation:

thats the answer

8 0
3 years ago
Help please!! Stuck on this question!! ∠ABC is adjacent to ∠CBD. If the m∠ABC=4x+23, m∠CBD=6x+7, and m∠ABD=130°, what is the mea
Ulleksa [173]

Answer:

63 degrees

Step-by-step explanation:

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3 0
3 years ago
Productivity: My candy bar company made 100 million bars last year, sold for $1 each. I also paid [L] people last year, with an
Murljashka [212]

Answer:

5

Step-by-step explanation:

Productivity: My candy bar company made 100 million bars last year, sold for $1 each. I also paid [L] people last year, with an average salary of $100K last year. I have overhead cost of $10M. What was my TOTAL productivity (no units, rounded to 2 decimal places)?

Solution:

Total productivity is the average of labour and capital productivity weighted and adjusted to price fluctuations. It is the ratio of total output to the total input. The total productivity is given by the formula:

Total productivity = total output / total input

Total output = Revenue = number of bars sold * price per bar

Total output = 100 million * $1 = $100 million

Total input = Total salary + overhead cost

Total salary = number of people * average salary = 100 *$100000 = $10 million

overhead cost = $10 million

Total input = $10 million + $10 million = $20 million

Total productivity = total output / total input = $100 million / $20 million

Total productivity = 5

7 0
2 years ago
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