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Harrizon [31]
3 years ago
6

4 divided by 2 over 5

Mathematics
2 answers:
Wittaler [7]3 years ago
8 0
.4
since 4/2 is 2, 2 divided by 5 is .4
umka2103 [35]3 years ago
6 0
Hey you can use Mathaway it's really it will solve anything you want
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Brad swims 4 laps every 6min how long does it take him to swim 14 laps?
Nadya [2.5K]
I don’t know the answer
5 0
3 years ago
Hurry i need help!!
diamong [38]

Answer:

Take rabbit across

go back

Take fox across and go back across with rabbit

take carrots across and leave them with fox

go back and get rabbit

Step-by-step explanation:

4 0
3 years ago
Which two integers is square root -21 between?
Pachacha [2.7K]

Answer:

None

Step-by-step explanation:

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7 0
2 years ago
Read 2 more answers
Find the area of the missing piece of the polygon and explain your process for finding it:
Otrada [13]

Answer:

42

Step-by-step explanation:

In short, the sum of the opposite areas are equal.

x + 30 = 24 + 48

x = 42

To prove this, draw a line from each corner to the "center" where the four lines meet.  Along each side of the square are two triangles.  These triangles have the same base and the same height, and therefore have the same area.

If we say the triangles at the bottom have area a, the triangles on the left have area b, the triangles on top have area c, and the triangles on the right have area d, then we can write 4 equations:

a + b = x

b + c = 24

c + d = 30

a + d = 48

Adding the first and third equations:

a + b + c + d = x + 30

Adding the second and fourth equations:

a + b + c + d = 24 + 48

Therefore:

x + 30 = 24 + 48

x = 42

8 0
3 years ago
Solve for the following equation step by step and justify your steps when using an exponential property.
mel-nik [20]

x = \frac{5}{7}

using the ' laws of exponents '

• a^{m} × a^{n} = a^{( m + n )}

•  a^{m} ÷ a^{n} = a^{(m - n)}

• (a^{m})^n = a^{mn}

7^{2} × 7^{3} = 7^{5} ÷ 7^{3x} = 7^{5 - 3x}

7^{4x} = 7^{5 - 3x}

equating the exponents since bases on both sides are 7

4x = 5 - 3x

7x = 5 ⇒ x =\frac{5}{7}




4 0
3 years ago
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