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Whitepunk [10]
3 years ago
13

Y ≥ 12x – 1 y < 43x – 2

Mathematics
1 answer:
kondaur [170]3 years ago
5 0

Answer:

Step-by-step explanation:

The solution to this problem is a graph.

Graph y ≥ 12x - 1:  this is a straight solid line with slope 12 and y-intercept -1.  Due to the presence of the  ≥  operator, we must shade (darken) the entire area above this line.  

Next, graph y < 43x – 2 using a dashed line (due to the < operator):  The slope is 43 and the y-intercept is -2.

Determine where the two lines intersect, either from the graph or algebraically.  If algebraically:

equate the two givens:

y = 12x - 1 = y = 43x – 2, which simplifies first to:

12x - 1 = 43x - 2, and second to

1 = 43x - 12x = 31x, which results in x = 1/31.

Substituting 1/31 for x in the first equation, we get:

y = 12(1/31) - 1, or y = 12/31 - 1, or y = 12/31 - 31/31, or y = 19/31

The two graphs intersect at (1/31, 19/31)

Shade (darken) the area beneath (under) the dashed line y < 43x – 2.

The solution is the area that you have darkened twice.

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Answer:

Rate of Pay is $11 per hour.

Step-by-step explanation:

Cost of 7\frac{1}{2}\:hr of work = $ 82.50

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7\frac{1}{2}=\frac{7\times2+1}{2}=\frac{15}{2}

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a_sh-v [17]

Answer:

Check the explanation

Step-by-step explanation:

(a)Let p be the smallest prime divisor of (n!)^2+1 if p<=n then p|n! Hence p can not divide (n!)^2+1. Hence p>n

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