The width of a rectangle is 3 cm less than twice its length. If the perimeter of the
rectangle is 18 cm, find its dimensions.

Answers

Answer 1

Answer:

length=4

Width=2×4-3=5

Step-by-step explanation:

Let length =X

Width =2x-3

Perimeter=18

2(length+width)=18

2(X+2x-3)=18

3x-3=9

3x=12

X=12/3

X=4

So length=4

Width=2×4-3=5

Answer 2

Answer:

[tex] \huge{ \boxed{ \bold{ \sf{Length \: = \: 4 \: cm}}}}[/tex]

[tex] \huge{ \boxed{ \bold{ \sf{Width = 5 \: cm}}}}[/tex]

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Question :

The width of a rectangle is 3 cm less than twice its length. If the perimeter of the rectangle is 18 cm, find its dimensions.

Step - by - step explanation :

✒ First, Let length of a rectangle be ' L '.

✑ Given :

Width of a rectangle = 3 cm less than twice its length ( i.e W = 2L - 3 )Perimeter of a rectangle ( P ) = 18 cm

Finding value of length of a rectangle :

[tex] \boxed{ \sf{Perimeter \: of \: a \: rectangle = 2(L + W)}}[/tex]

Plug the values and solve for L

➺ [tex] \sf{18 = 2(L+ 2L - 3)}[/tex]

➺ [tex] \sf{18 = 2(3L - 3)}[/tex] { Add : 2L and L )

➺ [tex] \sf{18 = 6L - 6}[/tex] { Distribute 2 through the parentheses )

➺ [tex] \sf{6L - 6 = 18}[/tex] { Swipe the sides of the equation }

➺ [tex] \sf{6L = 18 + 6}[/tex] { Move 6 to right hand side and change it's sign )

➺ [tex] \sf{6L = 24}[/tex] { Add the numbers : 18 and 6 }

➺ [tex] \sf{ \frac{6L}{6} = \frac{24}{6}} [/tex] { Divide both sides by 6 }

➺ [tex] \boxed{ \sf{length \: ( \: L\: ) \: = \: 4 \: cm}}[/tex]

☞ Now , Evaluating 2L - 3 when L = 4 cm in order to find the value of width of a rectangle : [tex] \sf{2l - 3}[/tex]

➺ [tex] \sf{2 \times 4 - 3}[/tex] { Plug the value of length }

➺ [tex] \sf{8 - 3}[/tex] { Multiply 2 by 4 }

➺ [tex] \boxed{ \sf{Width \: ( \: W \: ) \: = \: 5 \: cm}}[/tex]

Hence ,

Length of a rectangle ( L ) = 4 cmWidth of a rectangle ( W ) = 5 cm

Hope I helped!

Have a wonderful day! ツ

~TheAnimeGirl ♡

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The Width Of A Rectangle Is 3 Cm Less Than Twice Its Length. If The Perimeter Of Therectangle Is 18 Cm,

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Answers

Answer:

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Step-by-step explanation:

PLS GIVE BRAINLIEST

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= 27/136

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Answers

Answer:

as

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so

[tex]y'=0[/tex]

Step-by-step explanation:

Given the function

[tex]y=\:1-x+\frac{x^2}{1}+x-x^2[/tex]

Taking derivative

[tex]\frac{d}{dx}\left(1-x+\frac{x^2}{1}+x-x^2\right)[/tex]

[tex]\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'[/tex]

[tex]=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{1}\right)+\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(x^2\right)[/tex]

as

[tex]\frac{d}{dx}\left(1\right)=0[/tex]         ∵ [tex]\mathrm{Derivative\:of\:a\:constant}:\quad \frac{d}{dx}\left(a\right)=0[/tex]

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[tex]\frac{d}{dx}\left(\frac{x^2}{1}\right)=2x[/tex]      ∵ [tex]\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}[/tex]

[tex]\frac{d}{dx}\left(x\right)=1[/tex]          ∵ [tex]\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1[/tex]

[tex]\frac{d}{dx}\left(\frac{x^2}{1}\right)=2x[/tex]      ∵ [tex]\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}[/tex]

substituting all the values in the expression      

[tex]=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{1}\right)+\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(x^2\right)[/tex]

[tex]=0-1+2x+1-2x[/tex]

[tex]=0[/tex]

Hence,

[tex]\frac{d}{dx}\left(1-x+\frac{x^2}{1}+x-x^2\right)=0[/tex]

Therefore,

[tex]y'=0[/tex]


What is the area, height, and base of this triangle?

Answers

This should be your answer

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Answer: 22 x 10 = 220
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