Answer:
3,750 cars.
Step-by-step explanation:
We are given that the equation:

Models the relationsip between <em>y</em>, the number of unfilled seats in the stadium, and <em>x</em>, the number of cars in the parking lot.
We want to determine the number of cars in the parking lot when there are no unfilled seats in the stadium.
When there are no unfilled seats in the stadium, <em>y</em> = 0. Thus:

Solve for <em>x</em>. Subtract 9000 from both sides:

Divide both sides by -2.4:

So, there will be 3,750 cars in the parking lot when there are no unfilled seats in the stadium.
Answer:
option 2 is correct
Step-by-step explanation:
Answer:
Step-by-step explanation:
Hi,
I am quite good at maths and will be happy to help you, not just in Algebra, but any other topics in maths
In the
direction we consider the
subintervals [0, 1] and [1, 2] (each with length 1), while in the
direction we consider the
subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of
are (1, 0), (2, 0), (1, 2), (2, 2).
Let
. The volume of the solid is approximately

###
More generally, the lower-right-corner Riemann sum over
and
subintervals would be

Then taking the limits as
and
leaves us with an exact volume of
.
Answer:
the answer is B
Step-by-step explanation: