To solve the given algebraic expressions, we will substitute the provided values for the variables and simplify step by step. Let's go through each problem:
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1) \( p^2 + m \); where \( m = 2 \), \( p = 1 \)
Substitute \( m = 2 \) and \( p = 1 \):
\[
p^2 + m = (1)^2 + 2 = 1 + 2 = 3
\]
Answer: \( \boxed{3} \)
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2) \( x - x^2 + 1 + y \); where \( x = 1 \), \( y = -1 \)
Substitute \( x = 1 \) and \( y = -1 \):
\[
x - x^2 + 1 + y = 1 - (1)^2 + 1 + (-1)
\]
Simplify:
\[
= 1 - 1 + 1 - 1 = 0
\]
Answer: \( \boxed{0} \)
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3) \( p(r + q) \); where \( p = 3 \), \( q = 6 \), \( r = 4 \)
Substitute \( p = 3 \), \( q = 6 \), and \( r = 4 \):
\[
p(r + q) = 3(4 + 6) = 3(10) = 30
\]
Answer: \( \boxed{30} \)
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4) \( y \cdot (x - x) \); where \( x = 6 \), \( y = 2 \)
Substitute \( x = 6 \) and \( y = 2 \):
\[
y \cdot (x - x) = 2 \cdot (6 - 6) = 2 \cdot 0 = 0
\]
Answer: \( \boxed{0} \)
---
5) \( p(p + m) \); where \( m = 1 \), \( p = 6 \)
Substitute \( m = 1 \) and \( p = 6 \):
\[
p(p + m) = 6(6 + 1) = 6(7) = 42
\]
Answer: \( \boxed{42} \)
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6) \( j^2 h \); where \( h = 2 \), \( j = 4 \)
Substitute \( h = 2 \) and \( j = 4 \):
\[
j^2 h = (4)^2 \cdot 2 = 16 \cdot 2 = 32
\]
Answer: \( \boxed{32} \)
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7) \( xy \div 6 \); where \( x = 5 \), \( y = 6 \)
Substitute \( x = 5 \) and \( y = 6 \):
\[
xy \div 6 = (5 \cdot 6) \div 6 = 30 \div 6 = 5
\]
Answer: \( \boxed{5} \)
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8) \( 2p - q \); where \( p = 4 \), \( q = -2 \)
Substitute \( p = 4 \) and \( q = -2 \):
\[
2p - q = 2(4) - (-2) = 8 + 2 = 10
\]
Answer: \( \boxed{10} \)
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9) \( y^2 - x \); where \( x = 1 \), \( y = 5 \)
Substitute \( x = 1 \) and \( y = 5 \):
\[
y^2 - x = (5)^2 - 1 = 25 - 1 = 24
\]
Answer: \( \boxed{24} \)
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10) \( 6 - (p - m) \); where \( m = 1 \), \( p = -5 \)
Substitute \( m = 1 \) and \( p = -5 \):
\[
6 - (p - m) = 6 - ((-5) - 1) = 6 - (-6) = 6 + 6 = 12
\]
Answer: \( \boxed{12} \)
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11) \( (x - y) \div 2 \); where \( x = 3 \), \( y = 1 \)
Substitute \( x = 3 \) and \( y = 1 \):
\[
(x - y) \div 2 = (3 - 1) \div 2 = 2 \div 2 = 1
\]
Answer: \( \boxed{1} \)
---
12) \( (x + y)^2 \); where \( x = 1 \), \( y = 1 \)
Substitute \( x = 1 \) and \( y = 1 \):
\[
(x + y)^2 = (1 + 1)^2 = 2^2 = 4
\]
Answer: \( \boxed{4} \)
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Final Answers:
1. \( \boxed{3} \)
2. \( \boxed{0} \)
3. \( \boxed{30} \)
4. \( \boxed{0} \)
5. \( \boxed{42} \)
6. \( \boxed{32} \)
7. \( \boxed{5} \)
8. \( \boxed{10} \)
9. \( \boxed{24} \)
10. \( \boxed{12} \)
11. \( \boxed{1} \)
12. \( \boxed{4} \)
Boxed Final Answer:
\[
\boxed{3, 0, 30, 0, 42, 32, 5, 10, 24, 12, 1, 4}
\]
Parent Tip: Review the logic above to help your child master the concept of 6th grade algebra equations worksheet.
