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.
The value of the cosine ratio cos(L) is 5/13
<h3>How to determine the cosine ratio?</h3>
The complete question is added as an attachment
Start by calculating the hypotenuse (h) using
h^2 = 5^2 + 12^2
Evaluate the exponent
h^2 = 25 + 144
Evaluate the sum
h^2 = 169
Evaluate the exponent of both sides
h = 13
The cosine ratio is then calculated as:
cos(L) = KL/h
This gives
cos(L) =5/13
Hence, the value of the cosine ratio cos(L) is 5/13
Read more about right triangles at:
brainly.com/question/2437195
#SPJ1
The answer is B.C. =8.1cm
The working is on the picture
Let's represent the number of hours with the variable 'h' (or whatever letter you want it to be):
265 + 48h = 553
Substract both sides by 265:
48h = 288
Divide:
h = 6
6 hours of labor was spent
To get our answer, we just have to find the least common multiple (LCM) of 8 and 11. In this case, it's 88, because that's the smallest number that 8 and 11 are both factors of.
Hope this helped! :D