Y=8c+0.75t
Y represents total amount for a pizza with toppings
g (f(-3)) = -6 is your answer
Let <em>f(x)</em> be the sum of the geometric series,

for |<em>x</em>| < 1. Then taking the derivative gives the desired sum,

Answer:
2 hours and 30 minutes to fill it up
Step-by-step explanation:
Answer:
x≥2
Step-by-step explanation:
First, write out the equation as you have it:

Then, add
to both sides:

Next, subtract
from both sides:

Finally, divide both sides by
:

or

Therefore: the answer to this inequality/equation is: x≥2