Question about Dmm Conservatory 16 Foot x 8 Foot x 9 Foot 6 Inches

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If a rectangle has a width of 2376 mm and a length of 3555 mm what is the hypotenuuse

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4276 mm 

Posted on Oct 04, 2009

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The length, l, of a rectangle is 5/2 its width, w. The length, l, of the rectangle is 10 inches. What is the perimeter of the rectangle? What Iis the area of the rectangle, in sauare inches?


We are given the following data:

length = 5/2 * width
length = 10 inWe know through the transitive

property of equality that the following must be true:

5/2 * width = 10 in

We then solve this equation algebraically for width:

width = (10 in) * 2/5
width = 4 in

The perimeter of the rectangle is given by substituting our known values of length and width into the following general equation:

Perimeter = 2 * length + 2 * width
Perimeter = 2(10 in) + 2(4 in) = 28 in

In a similar fashion, we substitute these known values into the general equation for the area of the rectangle to solve for the same:

Area = length * width
Area = (10 in)(4 in)=40 in^2

Nov 29, 2016 | The Computers & Internet

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Suppose that the width of a certain rectangle is 5 inches more than one-fourth of its length. The perimeter of the rectangle is 50 inches. Find the length and width of the rectangle.


Ok
2L + 2W = 50 (i)
W = L/4 + 5 (ii)

So 4W = L + 20 from (ii) and
4W = 100 - 4L from (i)
L + 20 = 100 - 4L
5L = 80
L = 16
W = 9 from (ii)

check
(i) 2*16 + 2*9 = 50
(ii) 16/4 + 5 = 9 = W

Oct 10, 2014 | Computers & Internet

1 Answer

The perimeter of a rectangle is 600 yards. what are the dimensions of the rectangle if the length is 20 yards more then the width


160 by 140

The perimeter is 600, so 2w+2l=600 where w is the width and l is the length.
Divide both sides by 2: w+l=300
The length is 20 more than the width: l=w+20
Substituting in the previous equation: w+(w+20)=300
Collecting terms: 2w+20=300
Subtract 20 from both sides: 2w=280
Divide by 2: w=140
Thus the width is 140. Substituting into the equation for length: l=140+20
Simplifying: l=160
The width is 140 and the length is 160

Sep 15, 2014 | MathAid Algebra II

1 Answer

X(2x+3)=90


Solve the resulting quadratic equation
2X^2+3X-90=0
Discriminant: (3)^2-4(2)(-90)=729=(27)^2
Two roots
X_1=(1/4)*(-3+27)=6
X-2=(1/4)*(-3-27) =-(15/2), negative
Since the width must be positive, reject the negative root and keep X_1=6
Width =6
Length=2(6)+3=15

Check 6*(15)=90. Checks OK

Apr 23, 2014 | Office Equipment & Supplies

1 Answer

Volume of rectangle formula


A rectangle is a two-dimensional figure and doesn't have a volume. It has an area equal to length times width. A rectangular prism has a volume equal to length times width times height.

Feb 06, 2014 | Office Equipment & Supplies

2 Answers

How do you solve a rectangle is eight inches longer than its width. find the dimensions of the rectangle if the perimeter is 60 cm


The rectangle is 11cm by 19cm.

The perimeter is 60, so the width and length must add to 30. The length is 8 more than the width, or width+width+8=30. Solve that for width, then you can calculate the length based on the width.

If this is homework, be sure to show your work.

Oct 11, 2013 | Lands Phones

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Please show all of the information and answers to: The plane was purchased new in 2006; therefore, x =0 represents the year 2006. X - Axis (horizontal) = years starting from 0 = 2006 and increasing b


Translate the English into Mathematics.
W=L-5 (in inches)
P=2(L+W)=2(L+L-5)=2(2L-5)
Use distirbutive property of multiplication with respect to addition to open up the parentheses (brackets)
P=4L-10.
Set P= 50 (inches), to get 4L-10=50
Solve for L: Do it'!
Find W= L (the one you just found) -5 =

Now, with W the value you just calculated
the new length is L'=-4+3W
and the new perimeter is P'=2(L'+W)

Now your turn to do some work.

Jan 02, 2012 | Office Equipment & Supplies

1 Answer

Area of rectangle is 4122cm2 .width of rectangle is 3cm.how do i find the length of it


Area = length X width, and you know the area and the width. So,

4122 sq cm = length X 3 cm

length = 4122 sq cm / 3 cm

length = 1374 cm

(That is one long, skinny rectangle, isn't it?)

Mar 02, 2011 | Office Equipment & Supplies

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