Question
5. Match each β with a π .
A card sort is on the screen. To start dragging a card, Tab to it and press M to mark it. Next, Tab to the card where you want to drop, and press J to join the two cards together. To unjoin the selected card from a group, press U.
3xβ2 π
1
xβ β
3
x4β β
y3
x2β β
x2yβ3 π
x0 π
xβ2y3 π
xβ1 π
xβ3 π
1
x3β β
3xβ4 π
3
x2β β
1 β
To attach cards, drag it on TOP of each other until you see them CONNECT.
To detach, drag the BOTTOM card away.
x2
y3β
A card sort is on the screen. To start dragging a card, Tab to it and press M to mark it. Next, Tab to the card where you want to drop, and press J to join the two cards together. To unjoin the selected card from a group, press U.
3xβ2 π
1
xβ β
3
x4β β
y3
x2β β
x2yβ3 π
x0 π
xβ2y3 π
xβ1 π
xβ3 π
1
x3β β
3xβ4 π
3
x2β β
1 β
To attach cards, drag it on TOP of each other until you see them CONNECT.
To detach, drag the BOTTOM card away.
x2
y3β
Answers
Answered by
GPT-4o mini
To match each starred cards (β) with diamond cards (π), we need to look for equivalent algebraic expressions based on exponents of \( x \) and \( y \).
Hereβs how we can match them:
1. **\( \frac{1}{x} \) (β)** can be matched with **\( x^{-1} \) (π)** because \( \frac{1}{x} = x^{-1} \).
2. **\( \frac{3}{x^4} \) (β)** can be matched with **\( 3x^{-4} \) (π)** for the same reason: \( \frac{3}{x^4} = 3x^{-4} \).
3. **\( y^3x^2 \) (β)** can be matched with **\( x^2y^3 \) (π)** because these two expressions are equivalent regardless of the order (commutative property of multiplication).
4. **\( x^{-2}y^3 \) (β)** can be matched with **\( x^{-2}y^3 \) (π)** as they are the same expression.
5. **\( \frac{1}{x^3} \) (β)** can be matched with **\( x^{-3} \) (π)** since \( \frac{1}{x^3} = x^{-3} \).
6. **\( 1 \) (β)** remains unmatched as there is no specific partner among diamonds that is equivalent.
7. **\( 3x^2 \) (β)** can also remain unmatched as there is no specific partner among diamonds that is equivalent.
Therefore, the matches are:
1. \( 1/x \) β matches (x^{-1}) π
2. \( 3/x^4 \) β matches (3x^{-4}) π
3. \( y^3*x^2 \) β matches (x^2*y^3) π
4. \( x^{-2}y^3 \) β matches (x^{-2}y^3) π
5. \( 1/x^3 \) β matches (x^{-3}) π
Remaining unmatched cards are:
- \( 1 \) β
- \( 3x^2 \) β
You can now proceed with these matches as per the directions provided on the screen for the drag and drop action.
Hereβs how we can match them:
1. **\( \frac{1}{x} \) (β)** can be matched with **\( x^{-1} \) (π)** because \( \frac{1}{x} = x^{-1} \).
2. **\( \frac{3}{x^4} \) (β)** can be matched with **\( 3x^{-4} \) (π)** for the same reason: \( \frac{3}{x^4} = 3x^{-4} \).
3. **\( y^3x^2 \) (β)** can be matched with **\( x^2y^3 \) (π)** because these two expressions are equivalent regardless of the order (commutative property of multiplication).
4. **\( x^{-2}y^3 \) (β)** can be matched with **\( x^{-2}y^3 \) (π)** as they are the same expression.
5. **\( \frac{1}{x^3} \) (β)** can be matched with **\( x^{-3} \) (π)** since \( \frac{1}{x^3} = x^{-3} \).
6. **\( 1 \) (β)** remains unmatched as there is no specific partner among diamonds that is equivalent.
7. **\( 3x^2 \) (β)** can also remain unmatched as there is no specific partner among diamonds that is equivalent.
Therefore, the matches are:
1. \( 1/x \) β matches (x^{-1}) π
2. \( 3/x^4 \) β matches (3x^{-4}) π
3. \( y^3*x^2 \) β matches (x^2*y^3) π
4. \( x^{-2}y^3 \) β matches (x^{-2}y^3) π
5. \( 1/x^3 \) β matches (x^{-3}) π
Remaining unmatched cards are:
- \( 1 \) β
- \( 3x^2 \) β
You can now proceed with these matches as per the directions provided on the screen for the drag and drop action.