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Answer :
Sure, let's solve the problem step-by-step to find the dimensions of the rectangle.
We are given that the area of the rectangle [tex]\( A \)[/tex] is represented by the expression [tex]\( 24x^6y^{15} \)[/tex]. We need to find which pair of dimensions (length and width) multiplies together to give this area.
Let's evaluate each given pair to see if their product matches [tex]\( 24x^6y^{15} \)[/tex].
1. First Pair:
- Length: [tex]\( 2x^5y^8 \)[/tex]
- Width: [tex]\( 12xy^7 \)[/tex]
To find the area with these dimensions, multiply them together:
[tex]\[
(2x^5y^8) \times (12xy^7) = 2 \times 12 \times x^{5+1} \times y^{8+7} = 24x^6y^{15}
\][/tex]
The product is [tex]\( 24x^6y^{15} \)[/tex], which is exactly the given area.
2. Second Pair:
- Length: [tex]\( 6x^2y^3 \)[/tex]
- Width: [tex]\( 4x^3y^5 \)[/tex]
To find the area with these dimensions, multiply them together:
[tex]\[
(6x^2y^3) \times (4x^3y^5) = 6 \times 4 \times x^{2+3} \times y^{3+5} = 24x^5y^8
\][/tex]
The product is [tex]\( 24x^5y^8 \)[/tex], which is not the given area.
3. Third Pair:
- Length: [tex]\( 10x^6y^{15} \)[/tex]
- Width: [tex]\( 14x^6y^{15} \)[/tex]
To find the area with these dimensions, multiply them together:
[tex]\[
(10x^6y^{15}) \times (14x^6y^{15}) = 10 \times 14 \times x^{6+6} \times y^{15+15} = 140x^{12}y^{30}
\][/tex]
The product is [tex]\( 140x^{12}y^{30} \)[/tex], which is not the given area.
4. Fourth Pair:
- Length: [tex]\( 9x^4y^{11} \)[/tex]
- Width: [tex]\( 12x^2y^4 \)[/tex]
To find the area with these dimensions, multiply them together:
[tex]\[
(9x^4y^{11}) \times (12x^2y^4) = 9 \times 12 \times x^{4+2} \times y^{11+4} = 108x^6y^{15}
\][/tex]
The product is [tex]\( 108x^6y^{15} \)[/tex], which is not the given area.
Based on the evaluations, the pair of dimensions that multiply together to give the area [tex]\( 24x^6y^{15} \)[/tex] is [tex]\( 2x^5y^8 \)[/tex] and [tex]\( 12xy^7 \)[/tex].
So, the dimensions of the rectangle are [tex]\( 2x^5y^8 \)[/tex] and [tex]\( 12xy^7 \)[/tex].
We are given that the area of the rectangle [tex]\( A \)[/tex] is represented by the expression [tex]\( 24x^6y^{15} \)[/tex]. We need to find which pair of dimensions (length and width) multiplies together to give this area.
Let's evaluate each given pair to see if their product matches [tex]\( 24x^6y^{15} \)[/tex].
1. First Pair:
- Length: [tex]\( 2x^5y^8 \)[/tex]
- Width: [tex]\( 12xy^7 \)[/tex]
To find the area with these dimensions, multiply them together:
[tex]\[
(2x^5y^8) \times (12xy^7) = 2 \times 12 \times x^{5+1} \times y^{8+7} = 24x^6y^{15}
\][/tex]
The product is [tex]\( 24x^6y^{15} \)[/tex], which is exactly the given area.
2. Second Pair:
- Length: [tex]\( 6x^2y^3 \)[/tex]
- Width: [tex]\( 4x^3y^5 \)[/tex]
To find the area with these dimensions, multiply them together:
[tex]\[
(6x^2y^3) \times (4x^3y^5) = 6 \times 4 \times x^{2+3} \times y^{3+5} = 24x^5y^8
\][/tex]
The product is [tex]\( 24x^5y^8 \)[/tex], which is not the given area.
3. Third Pair:
- Length: [tex]\( 10x^6y^{15} \)[/tex]
- Width: [tex]\( 14x^6y^{15} \)[/tex]
To find the area with these dimensions, multiply them together:
[tex]\[
(10x^6y^{15}) \times (14x^6y^{15}) = 10 \times 14 \times x^{6+6} \times y^{15+15} = 140x^{12}y^{30}
\][/tex]
The product is [tex]\( 140x^{12}y^{30} \)[/tex], which is not the given area.
4. Fourth Pair:
- Length: [tex]\( 9x^4y^{11} \)[/tex]
- Width: [tex]\( 12x^2y^4 \)[/tex]
To find the area with these dimensions, multiply them together:
[tex]\[
(9x^4y^{11}) \times (12x^2y^4) = 9 \times 12 \times x^{4+2} \times y^{11+4} = 108x^6y^{15}
\][/tex]
The product is [tex]\( 108x^6y^{15} \)[/tex], which is not the given area.
Based on the evaluations, the pair of dimensions that multiply together to give the area [tex]\( 24x^6y^{15} \)[/tex] is [tex]\( 2x^5y^8 \)[/tex] and [tex]\( 12xy^7 \)[/tex].
So, the dimensions of the rectangle are [tex]\( 2x^5y^8 \)[/tex] and [tex]\( 12xy^7 \)[/tex].
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