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Black_prince [1.1K]
4 years ago
15

The length of a rectangular sandbox is

Mathematics
1 answer:
weeeeeb [17]4 years ago
7 0

Answer:

so the correct statement is;

The equation 2 ( 3 w − 6 ) + 2 w = 36 can be used to find the width of the rectangle, and the width is 6 feet

Step-by-step explanation:

Here, we want to select the statement which is true

Let the width be w

the length is 6 ft less than 3 times the width

The length will be (3w - 6) ft

Mathematically, the perimeter of a rectangle can be calculated using the formula;

2( l + b) = P

Thus;

2(3w -6 + w) = 36

divide both sides by 2

3w -6 + w = 36/2

3w -6 + w = 18

4w -6 = 18

4w = 18 + 6

4w = 24

w = 24/4

w = 6

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(4y^2-3y+2)+(5y+12y-4)-(13y^2-6y+20)
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Answer:

-(9 y^2 - 20 y + 22) or if you need y = 10/9 + (7 i sqrt(2))/9 or y = 10/9 - (7 i sqrt(2))/9

Step-by-step explanation:

Simplify the following:

-(13 y^2 - 6 y + 20) + 4 y^2 + 12 y + 5 y - 3 y - 4 + 2

Grouping like terms, -(13 y^2 - 6 y + 20) + 4 y^2 + 12 y + 5 y - 3 y - 4 + 2 = -(13 y^2 - 6 y + 20) + 4 y^2 + (-3 y + 5 y + 12 y) + (2 - 4):

-(13 y^2 - 6 y + 20) + 4 y^2 + (-3 y + 5 y + 12 y) + (2 - 4)

-3 y + 5 y + 12 y = 14 y:

-(13 y^2 - 6 y + 20) + 4 y^2 + 14 y + (2 - 4)

2 - 4 = -2:

-(13 y^2 - 6 y + 20) + 4 y^2 + 14 y + -2

-(13 y^2 - 6 y + 20) = -13 y^2 + 6 y - 20:

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Grouping like terms, 4 y^2 - 13 y^2 + 14 y + 6 y - 20 - 2 = (-13 y^2 + 4 y^2) + (6 y + 14 y) + (-20 - 2):

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4 y^2 - 13 y^2 = -9 y^2:

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Answer: -(9 y^2 - 20 y + 22)

______________________________________________

Solve for y:

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Divide both sides by -9:

y^2 - (20 y)/9 + 22/9 = 0

Subtract 22/9 from both sides:

y^2 - (20 y)/9 = -22/9

Add 100/81 to both sides:

y^2 - (20 y)/9 + 100/81 = -98/81

Write the left hand side as a square:

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Take the square root of both sides:

y - 10/9 = (7 i sqrt(2))/9 or y - 10/9 = -(7 i sqrt(2))/9

Add 10/9 to both sides:

y = 10/9 + (7 i sqrt(2))/9 or y - 10/9 = -(7 i sqrt(2))/9

Add 10/9 to both sides:

Answer: y = 10/9 + (7 i sqrt(2))/9 or y = 10/9 - (7 i sqrt(2))/9

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