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olga nikolaevna [1]
3 years ago
14

Daniel is printing out copies of a presentation. It takes 3 minutes to print a color copy and 1 minute to print a black-and-whit

e copy. He needs to print at least 6 copies and must have the copies completed in no more than 12 minutes.
If the solution region represents the number of color copies and black-and-white copies that Daniel can print, determine which graph represents the solution set to the system of inequalities representing this situation.

Mathematics
1 answer:
Effectus [21]3 years ago
6 0
Work the information to set inequalities that represent each condition or restriction.

2) Name the variables. 

c: number of color copies
b: number of black-and-white copies

3) Model each restriction:

i) <span>It takes 3 minutes to print a color copy and 1 minute to print a black-and-white copy.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He needs to print at least 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) And must have the copies completed in no more than 12 minutes ⇒</span>

3c + b ≤ 12
<span />

4) Additional restrictions are c ≥ 0, and b ≥ 0 (i.e. only positive values for the number of each kind of copies are acceptable)

5) This is how you graph that:

i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.

ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.

iii) since c ≥ 0 and b ≥ 0, the region is in the first quadrant.

iv) The final region is the intersection of the above mentioned shaded regions.

v) You can see such graph in the attached figure.

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Find the value of n, correct to 4 significant figures.​
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6.708

Step-by-step explanation:

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  • \implies y\:\alpha \:x^2......(1)

  • Y varies inversely as the square root of z (Given)
  • \implies y\:\alpha\: \frac{1}{\sqrt z}......(2)

  • Combining (1) & (2), we find:

  • y\:\alpha \:\frac{x^2}{\sqrt z}

  • \implies y=\frac{kx^2}{\sqrt z} (Where k is constant of proportionality).....(3)

  • Now, when y = 2, x = 3 and z = 4, we find the value of k i.e. constant.

  • 2=\frac{k(3)^2}{\sqrt 4}

  • \implies 2=\frac{k(9)}{2}

  • \implies k =\frac{4}{9}

  • Plugging the value of k in (3), we find:

  • y=\frac{4x^2}{9\sqrt z} ....(4)

  • Next, in equation (4), plug y = 5, x = n and z = 16 and obtain the value of n by solving it.

  • 5=\frac{4(n)^2}{9\sqrt {16}}

  • \implies 5=\frac{4(n)^2}{9(4)}

  • \implies 5=\frac{(n)^2}{9}

  • \implies 5(9)=(n)^2

  • \implies 45=(n)^2

  • \implies n=\sqrt{45}

  • \implies n=6.70820393

  • \implies n\approx 6.708
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