Mathematics, 23.04.2021 14:40 favre22fs

The length of a rectangular floor is 4 feet longer than its width w. The area of the floor is 525 ft^2. A) Write a quadratic equation in terms of w that represents the situation. B) What are the dimensions of the floor? Show your work.

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Answer from: nauticatyson9

x^21x+25x-525-0

x^21x+25x-525-0xx^2 - 3.7_) +5^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21

Answer from: Quest

trough to 50 re s thr bke nedsh le. therfore te dkgj jdjvhvfhv fe sjfv adxcc 60.

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Answer from: Quest

answer: x

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