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a_sh-v
5 days ago
6

lucas and erick are factoring the polynomial 12x3 – 6x2 8x – 4. lucas groups the polynomial (12x3 8x) (–6x2 – 4) to factor. eric

k groups the polynomial (12x3 – 6x2) (8x – 4) to factor. who correctly grouped the terms to factor? explain.
Mathematics
2 answers:
zzz [12.3K]5 days ago
8 0
Both students provided correct answers. Given: The polynomial is - To determine: Who factored the terms correctly? Lucas grouped the polynomial as - Next, solving Lucas’s polynomial should give us results. Erick grouped the polynomial as - Now, we will solve Erick’s polynomial. The factors that both Lucas and Erick arrived at are the same. Each grouping yields identical results. Consequently, both students are correct.
Leona [12.6K]5 days ago
5 0
Both Lucas and Erick are right since polynomials can be factored through various groupings. Each method leads to a shared binomial factor across the groups. By applying the distributive property, this shared binomial can be factored out. Ultimately, both groupings yield the identical two binomial factors.
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1 month ago
Which statement is true about the graphs of the two lines y = –6 and x =1/6 ?
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16 days ago
Read 2 more answers
You are given a rectangular sheet of cardboard that measures 11 in. by 8.5 in. (see the diagram below). A small square of the sa
AnnZ [12381]

Answer:

Terrence's

Step-by-step explanation:

The cut length of the square corresponds to the box height.

1. a

In Anya's approach: 8.5 - 1.5 = 7, 11 - 1.5 = 9.5; with the height being 1.5, the volume calculates as height x length x width = 1.5 x 9.5 x 7 = 99.75 square inches.

Terrence's method: 8.5 - 3 = 5.5, 11 - 3 = 8. Volume = 5.5 x 8 x 3 = 132 square inches. Clearly, 99.75 < 132 square inches, indicating Terrence's approach would yield a larger volume.

1. b

The dimensions of the box rely on the length, width, and height of the cardboard cutout, thus varying cutting measurements for the same cardboard size can produce different box dimensions.

2. Squares are cut from each corner, thus the combined size of the two squares on the cardboard must be less than the shorter cardboard side. The shorter side is 8.5 inches, which divided by 2 results in 4.25 inches, therefore squares cannot exceed this size. If squares are taken out measuring 4.25 inches, it will leave a strip 2.5 inches wide that cannot form a box.

To cut out squares of 5 inches, how you position them can cause overlap or extend beyond the cardboard. While 5 + 5 equals ten, making the longer side of 11 inches adequate, the 8.5 inches side allows zero room for the 10 inches.

7 0
1 month ago
This isosceles triangle has two sides of equal length, a, that are longer than the length of the base, b. The perimeter of the t
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B refers to the base of the triangle,
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The measurement labeled as 'a' is larger than 'b' since those equal sides are longer than the base. Given "one of the longer sides measures 6.3 cm," we assign a = 6.3.


Substitute 6.3 for each 'a' in the equation and solve for b:
2a + b = 15.7
2(6.3) + b = 15.7
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1 month ago
Solve x − 5y = 6 for x.
Inessa [12570]
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8 0
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