Okay. fist off don't think of them as 5 units each, but 1. so if you were to put it on graphing paper and box squares each time. you would be able to get 8 in for both sides to make a square. Area then would be 1600ft squared because 8 ×5 is 40 and 40×40 is 1600. then you calculate how many squares you used. 8×8=64. and 80-64=16 so 16 pieces are left over. then 9×9 (increasing the square) is 81. so 17 pieces to increase it.
Answer:
C.
Step-by-step explanation:
it is C. because when i worked this out, C is what i got
Formula: Y=kx or y=ax or Y/x=a
K=12 k is also the slope!
Slope=12
Y=6
So If you plug in these values you get:
6=12x
Answer:
X=0.5 or 1/5
The complete question in the attached figure
we know that
(see the attached figure n 2 to understand the problem)[the surface area of one prism]=2*[x*x]+2*[x*y]+2*[x*y]----> 2x²+4xy
[the surface area of the sculpture]=2*[5*x*y]+2*[3*x*x]+2*[3*x*y]--> 6x²+16xy
now
<span>JD says the surface area of the sculpture is 4 times the surface area of one prism
</span>[the surface area of the sculpture]=4*(2x²+4xy)---> 8x²+16xy
we compare the value that JD says with the real value
(8x²+16xy) > (6x²+16xy)
the value that JD says is <span>greater in comparison with the real value
</span>This is because <span>JD should also subtract the areas of eight hidden surfaces.
the answer is
</span>
JD should also subtract the areas of eight hidden surfaces<span>
</span>
I don’t think that’s possible. Unless the answer they’re looking for is in the decimals, 63 is only divisible by 1, 3, 7, 9, 21, and 63. Seeing as the formula for the area of a rectangle is a= lw, having something like that problem wouldn’t be possible. I’m most likely wrong, but I’m not completely sure. Please feel free to correct me!!