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Correct answer
Correct answer (there may be other correct orders, correct indentation required):
- def my_sum(first, second):
- return first + second
Correct answer (there may be other correct orders, correct indentation required):
- def my_sum(first, second):
- return first + second
Correct answer (there may be other correct orders):
- Assume $A \land B \land C$.
- Then $A$ is true.
- Then $C$ is true.
- Since $A$ and $C$, we know $A \land C$
Correct answer (there may be other correct orders):
- Assume $A \land B \land C$.
- Then $A$ is true.
- Then $C$ is true.
- Since $A$ and $C$, we know $A \land C$
Correct answer (there may be other correct orders, correct indentation required):
- def my_sum(first, second):
- sum = 0
- sum += second
- sum += first
- return sum
Incorrect Blocks (these blocks should not be included in your solution):
- sum + first
- sum + second
Correct answer (there may be other correct orders):
- 1
- 6
- 8
- 7
- 9
- 4
- 2
- 3
- 5
- 10
- 11
Correct answer (there may be other correct orders):
- A
- AA
Incorrect Blocks (these blocks should not be included in your solution):
- B
- C
- D
Correct answer (there may be other correct orders):
- first
- second
Correct answer (in the specified order):
- all or nothing first
- all or nothing second
- all or nothing third
Correct answer (in the specified order):
- first
- second
- third
Correct answer (there may be other correct orders):
- Let $a, b, c$ be arbitrary positive integers.
- Assume that $a|b$ and $b|c.$
- By definition, there exists $q_1$ such that $b=a\cdot q_1$ and there exists $q_2$ such that $c = b\cdot q_2.$
- We can see that $c = b\cdot q_2 = (a\cdot q_1)\cdot q_2 = a \cdot (q_1\cdot q_2).$
- By definition $a|c.$
Correct answer (there may be other correct orders):
- Some correct block before the distractor group
- Correct code
Incorrect Blocks (these blocks should not be included in your solution):
- Distractor 1
- Distractor 2
Correct answer (there may be other correct orders):
- Some correct block before the distractor group
- Correct code
Incorrect Blocks (these blocks should not be included in your solution):
- Distractor 1
- Distractor 2
Correct answer (there may be other correct orders):
- Some correct block before the distractor group
- Correct code 1
- Some correct block between the distractor group
- Correct code 2
- Some correct block after the distractor group
Incorrect Blocks (these blocks should not be included in your solution):
- Distractor 2 for correct code 1
- Distractor 1 for correct code 1
- Distractor 1 for correct code 2
- Distractor 2 for correct code 2
Correct answer (there may be other correct orders):
- Base Case
- $n=1: 2^1 \gt 1$
- Induction Hypothesis
- Assume $2^k \gt k$ for some integer $1 \leq k \lt n$
- Inductive Step
- Let $n$ be an arbitrary integer.
- $2^n = 2 \cdot 2^{n-1}$
- $2^n \gt 2(n-1)$ by IH
- Since $2(n-1) \gt 2n-n$, then $2^n \gt n$
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Question
QID:
Title:
Element pl-order-blocks: Dragging blocks to form the solution of a problem
Variant
Started at:
2026-08-04 22:43:05 (CDT)
Duration:
0s
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