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zvonat [6]
3 years ago
9

Theorem: For any given area A , the rectangle that has the least perimeter is a square

Mathematics
1 answer:
xeze [42]3 years ago
7 0

Geometry ahh I wish I could help. :'(

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PLZ DO EXACTLY AS STATED, WILL GIVE BRAILIEST AND 30 POINTS!!
dsp73

Answer:

bro just did this on a TGA

Step-by-step explanation:

A. no solutions and a contradiction

B. infinate solutions and an identity

C. one solution and niether

3 0
3 years ago
Solve for x: 3x − 2 > 5x + 10. (5 points)
velikii [3]

Answer:

Option 2.. x > -6

Step-by-step explanation:

3x -5x - 2 > 5x -5x + 10

-2x - 2 + 2 > 10 + 2

-2x/-2 > 12/-2

<u>x > -6</u>

5 0
3 years ago
What is two hundred ninety one divided by two?
Diano4ka-milaya [45]
<span>two hundred ninety one divided by two = 291/2 = 145.5

</span>
4 0
3 years ago
Read 2 more answers
I think I’m missing something
zepelin [54]

Answer:

Step-by-step explanation:

no ur good u got all of them

4 0
3 years ago
Read 2 more answers
n an orchid show, seven orchids are to be placed along one side of the greenhouse. of the orchids, we see that the number of lin
NNADVOKAT [17]

Answer:

35

Step-by-step explanation:

7 orchids can be lined as 7!. This means that for the first orchid of the line, you can select 7 options. When you place the first orchid, for the second option you can select among 6 since 1 orchid has already been placed. Similarly, for the 3rd orchid of  the line, you have left 5 options. The sequence goes in this fashion and for 7 orchids, you have 7*6*5*4*3*2*1 possibilities. However, there is a restriction here. 3 of the orchids are white and 4 are levender. This means that it does not make a difference if we line 3 white orchids in an arbitrary order since it will seem the same from the outside. As a result, the options for lining the 7 orchids diminish. The reduction should eliminate the number of different lining within the same colors. Similar to 7! explanation above, 3 white orchids can be lined as 3! and 4 levender orchids can be lined as 4!. To eliminate these options, we  divide all options by the restrictions. The result is:

\frac{7!}{4!*3!} = 35. [(7*6*5*4*3*2*1/(4*3*2*1*3*2*1)]

3 0
3 years ago
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