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Allushta [10]
4 years ago
11

7/28 in simplest form

Mathematics
2 answers:
vampirchik [111]4 years ago
8 0
\frac { 7 }{ 28 }

28 is 4 times 7, so we will rewrite it like that.

28=4\cdot 7\\ \\ \frac { 7 }{ 28 } =\frac { 7 }{ 4\cdot 7 }

7 is 7 times 7, so we will rewrite it like that.

\frac { 7 }{ 28 } =\frac { 7 }{ 4\cdot 7 } =\frac { 1\cdot 7 }{ 4\cdot 7 } =\frac { 1 }{ 4 } \cdot \frac { 7 }{ 7 } =\frac { 1 }{ 4 } \cdot 1=\frac { 1 }{ 4 }

So the simplest form is,

\boxed { \frac { 1 }{ 4 }  }
Aneli [31]4 years ago
3 0
\sf \frac{7}{28}^{(7}= \frac{1}{4}=\boxed{\sf 0,25}
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Step-by-step explanation:

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3 years ago
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A circle is centered at J(3, 3) and has a radius of 12.
stealth61 [152]

Answer:

(-6,\, -5) is outside the circle of radius of 12 centered at (3,\, 3).

Step-by-step explanation:

Let J and r denote the center and the radius of this circle, respectively. Let F be a point in the plane.

Let d(J,\, F) denote the Euclidean distance between point J and point F.

In other words, if J is at (x_j,\, y_j) while F is at (x_f,\, y_f), then \displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}.

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Point F would be on this circle if d(J,\, F) = r. (In other words, the distance between F\! and the center of this circle is exactly equal to the radius of this circle.)

Point F would be outside this circle if d(J,\, F) > r. (In other words, the distance between F\! and the center of this circle exceeds the radius of this circle.)

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5 0
3 years ago
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Ann [662]

Answer:

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Studentka2010 [4]
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