Answer:
The correct option is c: the area of the 6-inch pizza is 1/4 the area of the 12-inch pizza.
Step-by-step explanation:
The area (A) of each pizza is given by:

Where:
r: is the radius
For the 12-inch pizza we have:

And for the 6-inch pizza we have:

The ratio between A₆ and A₁₂ is:

Hence:

Therefore, the correct option is c: the area of the 6-inch pizza is 1/4 the area of the 12-inch pizza.
I hope it helps you!
For this case we have the following expression:
414x - 812y - 414x
Rewriting we have:
414x - 414x - 812y
We group similar terms:
(414x - 414x) - 812y
Adding we have:
0 - 812y
- 812y
Answer:
The sum of a number and its opposite is 0, and the sum of any number and 0 is the number itself.
Option c would be correct
Part A
<h3>Answer:
h^2 + 4h</h3>
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Explanation:
We multiply the length and height to get the area
area = (length)*(height)
area = (h+4)*(h)
area = h(h+4)
area = h^2 + 4h .... apply the distributive property
The units for the area are in square inches.
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Part B
<h3>Answer:
h^2 + 16h + 60</h3>
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Explanation:
If we add a 3 inch frame along the border, then we're adding two copies of 3 inches along the bottom side. The h+4 along the bottom updates to h+4+3+3 = h+10 along the bottom.
Similarly, along the vertical side we'd have the h go to h+3+3 = h+6
The old rectangle that was h by h+4 is now h+6 by h+10
Multiply these expressions to find the area
area = length*width
area = (h+6)(h+10)
area = x(h+10) ..... replace h+6 with x
area = xh + 10x .... distribute
area = h( x ) + 10( x )
area = h( h+6 ) + 10( h+6 ) .... plug in x = h+6
area = h^2+6h + 10h+60 .... distribute again twice more
area = h^2 + 16h + 60
You can also use the box method or the FOIL rule as alternative routes to find the area.
The units for the area are in square inches.
Check the picture below.
something worth noticing 
so, we're really graphing x+2, with a hole at x = 3, however, when x = 3, we know that f(x) = 5, but but but, when x = 3, x+2 = 5, so we end up with a continuous line all the way, x ∈ ℝ, because the "hole" from the first subfunction, gets closed off by the second subfunction in the piece-wise.