How to put 1.3, 1 1/3, 1.34 in order from least to greatest

Mathematics

2 answers:

Natasha_Volkova [10]4 years ago

7 0

11/3, 1.3, 1.34 Is the correct way to put it. PLZ give me the Brainliest answer. :)

77julia77 [94]4 years ago

5 0

The answer is, 1.3, 1 1/3, 1.34.
1.3 = 1.30
1 1/3= 1 x 3 =3 + 1 = 4/3= 4 divided by 3= 1.33
1.34=1.34
So, it'd be 1.30 , 1 1/3 , 1.34.

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The expression m divided by 4 is the distance each person runs in a relay race that is m miles how far does each person run in a

alexira [117]

Answer: <em>Each person runs 3 miles</em>

Step-by-step explanation:

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This would make your equation 12/4, which is equal to 3

5 0

3 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!