Recent content by halvizo1031

1. Proportions math problem

reason i say quadratic is because if a 10cm is $5, and a 20cm take four times the amount of raw materials, then it should cost four times the amount of a 10cm pizza. thus, it would cost$20.
2. Proportions math problem

it seems to me that it would be quadratic.
3. Proportions math problem

well as we doubled the diameter, the raw materials quadrupled.
4. Proportions math problem

now the second part of this question says that suppose that the price of each pizza will be directly proportional to the amount of raw materials you use. if you were to model your pricing structure as "price as a function of diameter", then would you expect that model to be linear, quadratic, or...
5. Proportions math problem

Homework Statement T or F? a pizza with 20cm diameter will require approximately half of the raw materials of pizza of diameter 40cm. explain your answer. Homework Equations The Attempt at a Solution my thinking is this, if we take the area of both pizzas, then we get 100(pi)...
6. Solving equations

thanks! i appreciate everyone's help.
7. Solving equations

hold on, if i plug in 1 for x, wouldn't i get 0 = 1???
8. Solving equations

the way i see it is that for any value i plug in for x (except 1), the left hand side will always give me a complex answer which does not equal 1.
9. Solving equations

if x < 1 then i will also get a complex solution here....therefore, i will always have a complex solution to the equation for any value of x except 0?
10. Solving equations

it seems to me that stating that the equation will give a complex answer when x>1
11. Solving equations

i'm not quite sure what it is the question is looking for .....
12. Solving equations

thus, sqrt of 1-x will always give us a complex answer?
13. Solving equations

i need some time to think about the reasoning here....i'm not quite sure i see it.
14. Solving equations

i see....then, may I use the same reasoning for part (d)?
15. Solving equations

well my thinking for part ii is that since (x-3) is always smaller than (x), then we would have a small number minus a big number which would give us a negative solution.....thus, since the equation is equalled to 5, there must be no solution.