Will mark brainly if u show work !
1 answer:
You might be interested in
Response:
Detailed explanation:
The function features a horizontal asymptote which is
The function presents a horizontal asymptote which is
One function lacks a horizontal asymptote because the degree of its numerator surpasses that of the denominator,
The function displays a horizontal asymptote which is
Utilize the details to create inequalities that reflect each limitation or requirement.
2) Label the variables.
c: count of color copies
b: count of black-and-white copies
3) Define each constraint:
i) <span>Printing a color copy requires 3 minutes while a black-and-white copy takes 1 minute.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He must print a minimum of 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) Moreover, he must finish the prints within 12 minutes at most ⇒</span>
3c + b ≤ 12
<span />
4) Additional limits include c ≥ 0, and b ≥ 0 (meaning only non-negative counts are valid for each type of copy)
5) Here’s how to illustrate that:
i) For 3c + b ≤ 12: draw the line representing 3c + b = 12 and shade the area above and to the right of this line.
ii) For c + b ≥ 6: draw the line c + b = 6 and shade the area below and to the left of this line.
iii) Given that c ≥ 0 and b ≥ 0, the relevant region is located in the first quadrant.
iv) The concluding area is the overlap of the previously mentioned shaded regions.
v) The graph can be viewed in the attached figure.
2) Label the variables.
c: count of color copies
b: count of black-and-white copies
3) Define each constraint:
i) <span>Printing a color copy requires 3 minutes while a black-and-white copy takes 1 minute.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He must print a minimum of 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) Moreover, he must finish the prints within 12 minutes at most ⇒</span>
3c + b ≤ 12
<span />
4) Additional limits include c ≥ 0, and b ≥ 0 (meaning only non-negative counts are valid for each type of copy)
5) Here’s how to illustrate that:
i) For 3c + b ≤ 12: draw the line representing 3c + b = 12 and shade the area above and to the right of this line.
ii) For c + b ≥ 6: draw the line c + b = 6 and shade the area below and to the left of this line.
iii) Given that c ≥ 0 and b ≥ 0, the relevant region is located in the first quadrant.
iv) The concluding area is the overlap of the previously mentioned shaded regions.
v) The graph can be viewed in the attached figure.