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Ulleksa [173]
3 years ago
14

Find the greatest common factor of 11m^3 and 13m^2. Help!!!

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
4 0
The greatest common factor of 11m^3 and 13m^2 is m^2.  11 and 13 don't really have a greatest common factor (they're both prime numbers), and m^2 goes into each of them.
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Solve <br>60000000+63636733-7373 ​
Yuri [45]

Answer:

your answer is 123629360

Step-by-step explanation:

60000000+63636733-7373 =123629360

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2 years ago
Question 9
Ratling [72]

Answer:

Two sides of a triangle have lengths 10 and 15.

What must be true about the length of the third side?

Greater than 5 but less than 25

Less than 15

Less than 25

Step-by-step explanation:

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Two sides of a triangle have lengths 10 and 15.

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Greater than 5 but less than 25

Less than 15

Less than 25

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3 0
2 years ago
Two lines in a plane that never intersect are<br>called​
mel-nik [20]

Answer:

Two lines, both in the same plane, that never intersect are called parallel lines. Parallel lines remain the same distance apart at all times.

Step-by-step explanation:

parallel lines.

3 0
3 years ago
Read 2 more answers
HELP PLZ ALGEBRA 1 HW
nlexa [21]

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4 0
2 years ago
Lucia is wrapping packages that are in the shape of a triangular prism. The net of the prism is shown below: ( Help me answer th
solniwko [45]

Answer:

The total surface area of all 6 prisms is 6336 in^2.

Step-by-step explanation:

Let's find the surface area of ONE prism and then multiply that result by 6 to obtain the final answer.

One prism:

The area of the two 13 in by 26 in rectangular tabs is 2(13 in)(26 in), or 676 in^2 (subtotal);

The area of the two triangles of base 10 in and height 12 in is 2([1/2][10 in][12 in], or 120 in^2; and, finally,

The area of the 10 in by 26 in base is 260 in^2.

The total surface area of ONE prism is thus:

676 in^2 + 120 in^2 + 260 in^2, or 1056 in^2.

Now, because there are 6 of these prisms, multiply this last result by 6:

6(1056 in^2) = 6336 in^2.

The total surface area of all 6 prisms is 6336 in^2.

7 0
3 years ago
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