Asked by Anonymous
A rectangular poster has an area of 190in(squared). The height of the poster is 1in. less than twice its width. Find the dimensions of the poster.
Answers
Answered by
Anonymous
A = Area= 190 in^2
H = Hight
W = Width
H = 2W-1
A = W*H
A = W*(2W-1)
190 = 2W*W-1*W
190 = 2W^2-W
2W^2-W-190 = 0
The exact solutions of this equation are:
W=10
and
W= -19/2
Width can't be negative so:
W = 10 in
H = 2*W-1
H = 2*10-1
H = 20-1
H = 19 in
W = 10 in
H = 19 in
A = W*H
A = 10*19
A = 190 in^2
If you don't know solve quadratic equation, in google type:
quadratic equation online
When you see list of results click on:
Free Online Quadratic Equation Solver:Solve By Quadratic Formula
When page be open in rectangle type:
2W^2-W-190=0
and click option: solve it!
You will see solution step-by-step
H = Hight
W = Width
H = 2W-1
A = W*H
A = W*(2W-1)
190 = 2W*W-1*W
190 = 2W^2-W
2W^2-W-190 = 0
The exact solutions of this equation are:
W=10
and
W= -19/2
Width can't be negative so:
W = 10 in
H = 2*W-1
H = 2*10-1
H = 20-1
H = 19 in
W = 10 in
H = 19 in
A = W*H
A = 10*19
A = 190 in^2
If you don't know solve quadratic equation, in google type:
quadratic equation online
When you see list of results click on:
Free Online Quadratic Equation Solver:Solve By Quadratic Formula
When page be open in rectangle type:
2W^2-W-190=0
and click option: solve it!
You will see solution step-by-step
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.