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Mrac [35]
3 years ago
9

Find the area of a square if its sides measure 2 2/3m

Mathematics
1 answer:
olga55 [171]3 years ago
6 0

Answer:

7\dfrac{1}{9}\ m^2

Step-by-step explanation:

If the length of the square side is a units, then the area of the square is

A=a^2\ un^2.

In your case, the side length is

a=2\dfrac{2}{3}\ m,

then the area is

A=2\dfrac{2}{3}\cdot 2\dfrac{2}{3}\\ \\ \\A=\dfrac{2\cdot 3+2}{3}\cdot \dfrac{2\cdot 3+2}{3}\\ \\ \\A=\dfrac{8}{3}\cdot \dfrac{8}{3}\\ \\ \\A=\dfrac{64}{9}\\ \\ \\A=7\dfrac{1}{9}\ m^2

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Answer:

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Step-by-step explanation:

6 0
3 years ago
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Rodney is given twi equations:x-y=11 and 2x+y=19
lozanna [386]

Answer:

x= 10 and y= -1

Step-by-step explanation:

Rewrite your second equation Y= -2x+ 19 then plug it into the first equation.

x- (-2x + 19) = 11

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The plug that answer into the first equation and solve.

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3 0
3 years ago
Please Help this is Due in 1 hour and I really dont understand any of this. (Question in Image)
vaieri [72.5K]

Answer:

(-4, -1/2)

Step-by-step explanation:

to calculate the midpoints bt of the line use the ormula:

( \frac{x1 + x2}{2} \frac{y1 + y2}{2} )

Where (x1, x2) is one coordinate point and (y1, y2) is anither coordinate point. Any two points will work, but I chose A (-5,-4) and B (-3, 3).

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5 0
2 years ago
Order these numbers from least to greatest.
PSYCHO15rus [73]
<h3>Answer:  3.1691, 5.8, 5.802, 5.82</h3>

Explanation:

The smallest item is 3.1691 which is listed first. This is because 3 is smaller than 5.

Now to sort the values that start with 5.

Think of 5.82 as 5.820; think of 5.8 as 5.800

We can see that 820 is larger than 800, which means 5.820 is larger than 5.800; in short, 5.82 > 5.8

Through similar logic, we can see that 5.82 > 5.802 and it further means 5.82 is the largest item. The next largest is 5.802

The sub-list {5.802, 5.82, 5.8} sorts to {5.8, 5.802, 5.82}. These values are then written after the 3.1691 mentioned. This will produce the fully sorted list from smallest to largest.

4 0
2 years ago
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blagie [28]
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5 0
3 years ago
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