Thread: Just started class again, help with some review questions.

FIND THE NUMBER OF UNITS THAT GIVES BREAK-EVEN FOR THE PRODUCT. ROUND YOUR ANSWER TO THE NEAREST WHOLE UNIT. ANSWER IN A COMPLETE SENTENCE.

125) A manufacturer has total revenue given by the function R = 130x and has total cost given by C = 293,000 + 10x, where x is the number of units produced and sold.

SOLVE BY WRITING A THREE PART INEQUALITY THEN SOLVING IT. WRITE ANSWER IN A COMPLETE SENTENCE.

132) Assume that the mathematical model C(x) = 15x + 170 represents the cost C, in hundreds of dollars, for a certain manufacturer to produce x items. How many items x can be manufactured while keeping costs between $467,000 and $692,000.


SOLVE THE INEQUALITIES

144) -6z + 6 ≥ -7z + 4

145) -5 < (6-3x / 3) ≤ 7

What kind of help do you want? It would be better to show what you have done and where you are stumped. That way we would know better what you need.

In any case, for the first one, you just need to know that the "break even" point is where Revenue equals Cost: set R= C and solve for x.

For the second one, the key is " while keeping costs between $467,000 and $692,000." That says that . Solve that inequality for x.

For that and the others, you solve a linear inequality essentially just like you solve an equation. The one thing to be careful about is that if you multiply or divided both sides by a negative number, the "direction" of the inequality changes.

For example, you solve -2x< 4 by dividing both sides by -2: -2x/-2= x, 4/-2= -2 and because you divided by a negative number, "<" becomes ">". The solution is x> -2.

Thanks for the explanations. I'm seriously 100% stumped on the 2nd one, and the inequalities though. I understand the first one though.

This is all seriously new to me because I didn't have a math course senior year of high school, and I did bad in Trigonometry during Junior Year. So I basically need as much help I can get on the other ones. She's testing us on this information tomorrow so I'm trying to get an understanding for every problem.