An even function has a graph that is symmetric about the Y axis
and odd function is symmetric about the x axis
1st graph = odd
2nd graph = neither
3rd graph = even
That's a 33% increase.
I calculated this using the formula:

Where n = the new value (16 in your question), o = the old value (12 in your question) and the result is outputted as a percent increase. You can check that this is correct by finding 33% of 12, adding the result to 12, and checking that the result equals your "new" number.
Note that 33% is only an approximation as your question requires a number rounded to the nearest whole.
<em>429 cm²</em>
- Step-by-step explanation:
<em>A(blue) =</em>
<em>= 2×6cm×10cm + 2×6cm×8cm + 8cm×10cm + (8cm×10cm - 5cm×4cm)</em>
<em>= 120cm² + 96cm² + 80cm² + 60cm²</em>
<em>= 356 cm²</em>
<em>A(green) =</em>
<em>= 4cm×5cm + 2×5cm×5cm/2 + 4cm×7cm</em>
<em>= 20cm² + 25cm² + 28cm²</em>
<em>= 73 cm²</em>
<em>A(total) =</em>
<em>= A(blue) + A(green)</em>
<em>= 356 cm² + 73 cm²</em>
<em>= 429 cm²</em>
9514 1404 393
Answer:
(a) 6² +3² +1² +1² = 47
(b) 5² +4² +2² +1² +1² = 47
(c) 3³ +4² +2² = 47
Step-by-step explanation:
It can work reasonably well to start with the largest square less than the target number, repeating that approach for the remaining differences. When more squares than necessary are asked for, then the first square chosen may need to be the square of a number 1 less than the largest possible.
The approach where a cube is required can work the same way.
(a) floor(√47) = 6; floor(√(47 -6^2)) = 3; floor(√(47 -45)) = 1; floor(√(47-46)) = 1
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(b) floor(√47 -1) = 5; floor(√(47-25)) = 4; ...
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(c) floor(∛47) = 3; floor(√(47 -27)) = 4; floor(√(47 -43)) = 2