Which statements are correct about the two way frequency table? Check all that apply. The table displays quantitative data, the
2 answers:
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The answer is: "40 cm " .
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<u>Note</u>:
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;
Solve for "x" (in "cm" ) ;
→ 5x = 200 * 1 ;
→ 5x = 200 ;
Divide each side of the equation by "5" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ 5x / 5 = 200 / 5 ;
to get:
→ x = 40 .
___________________________________________________________
The answer is: " 40 cm " .
___________________________________________________________
_________________________________________________________
<u>Note</u>:
_________________________________________________________
Solve for "x" (in "cm" ) ;
→ 5x = 200 * 1 ;
→ 5x = 200 ;
Divide each side of the equation by "5" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ 5x / 5 = 200 / 5 ;
to get:
→ x = 40 .
___________________________________________________________
The answer is: " 40 cm " .
___________________________________________________________
<h3>Answer: -2w^2 + 25w = 25 or -2w^2 + 25w - 25 = 0</h3>
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Explanation:
Refer to the diagram below. The width is w. We have two opposite and parallel sides equal to this. The other two parallel congruent sides are L = 25-2w meters long. We start with the total amount of fencing, and then subtract off the two width values, so 25-w-w = 25-2w.
The area of the rectangle is
Area = length*width
Area = L*W
Area = (25-2w)*w
Area = 25w - 2w^2
Area = -2w^2 + 25w
Set this equal to the desired area (25 square meters) to get
-2w^2 + 25w = 25
and we can subtract 25 from both sides to get everything on one side
-2w^2 + 25w - 25 = 0
side note: The two approximate solutions of this equation are w = 1.0961 and w = 11.4039 (use the quadratic formula or a graphing calculator to find this)