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Liono4ka [1.6K]
4 years ago
8

The length of a rectangle is twice its width. Find its area, if its perimeter is 7 1/3 cm.

Mathematics
1 answer:
Masja [62]4 years ago
7 0
2 80/81 is the area of the rectangle
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List 5 fractions that are between 1/3 and 4/5
koban [17]

9514 1404 393

Answer:

  2/5, 7/15, 8/15, 3/5, 2/3

Step-by-step explanation:

If these fractions are expressed with a common denominator, that would be 3×5 = 15. Then the given fractions are 1/3 = 5/15, and 4/5 = 12/15. The numerators 5 and 12 differ by 7, so we can easily choose 5 fractions in that range:

  6/15 = 2/5

  7/15

  8/15

  9/15 = 3/5

  10/15 = 2/3

_____

<em>Alternate solutions</em>

There is no requirement for the fractions to be written any particular way or with any particular spacing. The limits in decimal are 1/3 = 0.3333...(repeating) and 4/5 = 0.8. We could choose the decimal fractions ...

  0.34, 0.40, 0.50, 0.60, 0.70

or

  0.41, 0.52, 0.63, 0.74, 0.79

7 0
3 years ago
In addition to the following closing costs listed below, the buyer pays a realtor commission that is 3.5% of the loan amount.
aleksklad [387]
Solving for the realtor commission we have:
$165,000 x 0.035 = $5775 (3.5% of the loan amount)

Solving for closing cost, we total all of the costs including the realtor commission:
$5775 + $280 + $476 + $675 + $200 + $118 + <span>$573 = $8,097.</span>
7 0
3 years ago
Which expression is equivalent to 5g+9+8g+4?
vfiekz [6]
13g+13 sorry if this is wrong
6 0
3 years ago
Read 2 more answers
A rectangle has a perimeter of 30 feet and an area of 50 square feet. What are the dimensions of the rectangle?
REY [17]
Perimeter (p) = 2×length (l) + 2×width (w)
p = 2l+2w
area (a) = l×w, so solve for one (I'll use l):
a = l \times w \\ l =  a \div w \\ p = 2l + 2w = 2(l + w)
since p = 30, and a = 50, substitute the "a÷w" in for l in the perimeter equation:
p = 2(l + w) = 2((a \div w) + w) \\  = 2((a \div w) + ( {w}^{2} \div w)) \\  p= 2((a +  {w}^{2}) \div w
Now plug in p and a values:
p= 2((a +  {w}^{2}) \div w) \\ 30 = 2((50 +  {w}^{2}) \div w)  \\ 30 \div 2 = (50 +  {w}^{2}) \div w \\ 15w = 50 +  {w}^{2}
15w = 50 +  {w}^{2}  \\  {w}^{2}  - 15w + 50 = 0 \\ (w - 5)(w - 10) = 0
therefore width can be either 5 or 10 (but not both), so let's plug in:
l = a÷w = 50÷5 = 10
So if w = 5, then l = 10

D) 5 feet, and 10 feet
3 0
4 years ago
I need help with this question
Mekhanik [1.2K]

Answer:

Nvm you got it then thats good

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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