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My name is Ann [436]
2 years ago
6

A rectangle has an area of 12y^2+21y^512 y 2 +21y 5 12, y, squared, plus, 21, y, start superscript, 5, end superscript.The width

of the rectangle is equal to the greatest common monomial factor of 12y^212y 2 12, y, squared and 21y^521 y 5 21, y, start superscript, 5, end superscript.What is the length and width of the rectangle?
Mathematics
1 answer:
umka21 [38]2 years ago
8 0

Answer:

<em>Width</em>: 3y²

<em>Length</em>: 4 + 7y³

Step-by-step explanation:

The area of the rectangle is the length multiplied by the width. For the area given: 12y² + 21y⁵

12y² = 3y²*4

21y⁵ = 3y²*7y³

So, the greatest common monomial factor is 3y², which is the width.

The area is 3y²*(4 + 7y³), the first term is the width, the other must be the length: 4 + 7y³.

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Please actually help me. ill give brainliest if you explain your answer.
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subtract 11 from both sides

Step-by-step explanation:

-3x+11=44

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x=-11

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Step-by-step explanation:

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Sarah has $20 saved. She gets $10 per week for her allowance, and she saves her allowance for the next 3 weeks. At the end of th
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2t

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3 years ago
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