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
There are no human answers yet.
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.
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