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poizon [28]
2 years ago
12

Which of the following conditions is not sufficient to prove that a quadrilateral is a parallelogram?

Mathematics
1 answer:
ella [17]2 years ago
8 0

Answer: D

Step-by-step explanation:

A counterexample is an isosceles trapezoid.

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The perimeter of a pool is 96 feet. If the length of the pool is 3 times its width, what is the width of the pool? Labeling the
k0ka [10]

Answer:

12 feets

Step-by-step explanation:

Let the area of the pool be Length×width

There will be 2lengths and 2widths as well so the perimeter of the pool will be the addition of all the sides of the pool i.e 2L + 2W = 96... (1)

And since the length of the pool is 3 times its width, we have

L = 3W... (2)

Substituting equation 2 into 1 to get the width 'W'

2(3W) + 2W = 96

6W + 2W = 96

8W = 96

W = 96/8

W = 12feets

The width of the pool is 12feets

4 0
3 years ago
1300 error interval to 2dp
Degger [83]

Answer:

Huh

Step-by-step explanation:

5 0
2 years ago
C) 4c°+4c°+16c°+90°=360°
skad [1K]

Answer:

4c + 4c + 16c + 90 = 360

Add the like terms

24c + 90 = 360

Subtract 90 when taking it to the other side of the equal

24c = 270

Divide 270 by 24 to get ur answer.

c = 11.25

4 0
2 years ago
Read 2 more answers
2*8*5=2*5*8 identify the property that is shown
ira [324]
The property shown is commutative multiplication.  This property says that the product of two factors is not affected by the order in which they're multiplied.  In this case, the product remains the same i.e.,  2*8*5 = 80 while 2*5*8 <em>also </em>equals 80.
7 0
3 years ago
Read 2 more answers
How many palindromes of length 5 can you form using letters with the following properties: they start with a consonant, and the
Zielflug [23.3K]

Answer:

  2100

Step-by-step explanation:

There are 21 consonants that can serve as the first and last letters. There are 5 vowels that can serve as the 2nd and 4th letters. There are 20 remaining consonants that can serve as the 3rd letter. (The same consonant cannot appear in all three places.)

So, the total number of 5-letter palindromes that start with a consonant and alternate with a vowel will be ...

  21×5×20 = 2100 . . . different palindromes

6 0
3 years ago
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