No.
f(-2) = -114, not zero.
(x-4) is the only rational factor of f(x).
f(-2) = -114, not zero.
(x-4) is the only rational factor of f(x).
Which equation would have real zero(s) corresponding to the x-intercept(s) of the graph below?
1 answer:
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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.
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.