Answer: The second term of the expression which is 0.25x is the variable cost which is the extra amount paid for being late.
The expression 2 + 0.25x represents the amount that Ruth paid for the book.
It should be noted that the fixed cost is represented by 2 while the 0.25 represents the variable cost which is the late fee. Also, the x represented the number of days for being late.
In conclusion, the second term of the expression represents the amount that was charged for being late.
Answer: This expression is equivalent to 3^x
Step-by-step explanation: In this question the important concept is to write 80 in exponential form and multiply the exponents.
=80^1/4x
=(3^4)^1/4x
=3^x
Answer:
Last choice
Step-by-step explanation: Subtract 11 and then divide by 3. A would be greater than -2 and last choice shows that
Answer:
Step-by-step explanation:
VC = (1/3)·pi·r2·h
VC = volume of a cone
r = radius of base of cone = 6 cm
h = height
VC = (1/3)·pi·62·h
VC = 12·pi·h
VS = (4/3)·pi·r3
VS = volume of a sphere
r = radius of base of sphere = 3 cm
VS = (4/3)·pi·33
VS = 36·pi
Since the volume of the cone and the sphere are equal:
VC = VS
12·pi·h = 36·pi
Solve for h.

<em>We should ISOLATE x</em>
<em />
<em>Find the Natural Log of Both Sides to Make the Left Side "y"</em>
<em />
<em>Now, FIND THE DERIVATIVE Using Chain Rule!!!</em>
<em />