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brilliants [131]

3 years ago

10

Which is greater 7/3 or 5/2? How do you know

2 answers:

OLga [1]3 years ago

6 0

In order to understand which one is greater, you need to make the denominator on both fractions equal.

7/3 (2)= 14/6
5/2 (3)= 15/6

Since 15/6 is greater than 14/6:

5/2 is the greater fraction.
Hope this helps!!!

labwork [276]3 years ago

5 0

You need to get the denominator at the same number first. The Lowest common denominator is 6. So multiple 7/3 by 2 and 5/2 by 3 to get 14/6 and 15/6. 15/6 is greater. 15/6 is 5/2, so the answer is 5/2. I hope this helps!

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Find the dimensions of an open rectangular box with a square base that holds 2000 cubic cm and is constructed with the least bui

Vesnalui [34]

<h3>The dimensions of the given rectangular box are:</h3><h3>L = 15.874 cm , B = 15.874 cm , H = 7.8937 cm</h3>

Step-by-step explanation:

Let us assume that the dimension of the square base = S x S

Let us assume the height of the rectangular base = H

So, the total area of the open rectangular box

= Area of the base + 4 x ( Area of the adjacent faces)

= S x S + 4 ( S x H) = S² + 4 SH ..... (1)

Also, Area of the box = S x S x H = S²H

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Substituting the value of H in (1), we get:

$A = S^2 + 4 SH = S^2 + 4 S(\frac{2000}{S^2}) = S^2 + (\frac{8000}{S})\\\implies A = S^2 + (\frac{8000}{S})$

Now, to minimize the area put :

$(\frac{dA}{dS} ) = 0 \implies 2S - \frac{8000}{S^2} = 0\\\implies S^3 = 4000\\\implies S = 15.874 \approx 16 cm$

Putting the value of S = 15.874 cm in the value of H , we get:

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Hence, the dimensions of the given rectangular box are:

L = 15.874 cm

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