Area of a rectangle of length L and width W is A = LW.
Here, A = 60 ft^2 = (12 ft)(2x-3 ft). Let's find the values of x for which the area =60 ft^2, and then the values of x for which the area is greater than 60 ft^2.
60 = 12(2x-3) becomes 5 = 2x - 3 after dividing both sides by 12.
Adding 3 to both sides of this last equation results in 8 = 2x, so x = 4.
Does (12 ft)(2[4 ft] - 3) = 60? Does (12 ft)(5 ft) = 60 sq ft? YES.
The area of this rectangle will be greater than 60 sq ft if x>4 ft. (answer)
Answer:
Step-by-step explanation:
48:44= 12:11
6, 7, 8, and 9. just plug the X into the equation
Answer: The dimensions are: " 1.5 mi. × ³⁄₁₀ mi. " .
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{ length = 1.5 mi. ; width = ³⁄₁₀ mi. } .
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Explanation:
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Area of a rectangle:
A = L * w ;
in which: A = Area = (9/20) mi.² ,
L = Length = ?
w = width = (1/5)*L = (L/5) = ?
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A = L * w ; we want to find the dimensions; that is, the values for
"Length (L)" and "width (w)" ;
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Plug in our given values:
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(9/20) mi.² = L * (L/5) ; in which: "w = L/5" ;
→ (9/20) = (L/1) * (L/5) = (L*L)/(1*5) = L² / 5 ;
↔ L² / 5 = 9/20 ;
→ (L² * ? / 5 * ?) = 9/20 ?
→ 20÷5 = 4 ; so; L² *4 = 9 ;
↔ 4 L² = 9 ;
→ Divide EACH side of the equation by "4" ;
→ (4 L²) / 4 = 9/4 ;
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to get: → L² = 9/4 ;
Take the POSITIVE square root of each side of the equation; to isolate "L" on one side of the equation; and to solve for "L" ;
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→ ⁺√(L²) = ⁺√(9/4) ;
→ L = (√9) / (√4) ;
→ L = 3/2 ;
→ w = L/5 = (3/2) ÷ 5 = 3/2 ÷ (5/1) = (3/2) * (1/5) = (3*1)/(2*5) = 3/10;
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Let us check our answers:
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(3/2 mi.) * (3/10 mi.) =? (9/20) mi.² ??
→ (3/2)mi. * (3/10)mi. = (3*3)/(2*10) mi.² = 9/20 mi.² ! Yes!
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So the dimensions are:
Length = (3/2) mi. ; write as: 1.5 mi.
width = ³⁄₁₀ mi.
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or; write as: " 1.5 mi. × ³⁄₁₀ mi. " .
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Answer:
A
Step-by-step explanation:
because it makes more sense
