The surface area of a cone is equal to (1/3)π(radius)^{2}x(height).

The volume of a cone is equal to (1/3)π(radius)^{2}x(height).

For a three-digit number where all three digits are the same, the number divided by the sum of the digits in the number is equal to 27.

For a three-digit number where all three digits are the same, the number divided by the sum of the digits in the number is equal to 37.

5^{7} = 7^{5}

5^{7} ≠ 7^{5}

1 is a prime number.

1 is not a prime number.

On a radian-based graph, sin(0)=sin(π).

On a radian-based graph, sin(0)≠sin(π).

1^{3}+2^{3}+ ... +n^{3} = (1+2+ ... + n)^{2}

1^{3}+2^{3}+ ... +n^{3} ≠ (1+2+ ... + n)^{2}

A parabola is always mirror-symmetrical.

It is possible for a parabola to be asymmetrical.

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+N) as the first two digits and 25 as the last two.

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

1^{3}+2^{3}+ ... +n^{3} = (1+2+ ... + n)^{2}

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+N) as the first two digits and 25 as the last two.

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

A parabola is always mirror-symmetrical.

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

'e' is a rational number.

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

The surface area of a cone is equal to (1/3)π(radius)^{2}x(height).

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

1 is a prime number.

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

On a radian-based graph, sin(0)=sin(π).

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

123456789+987654321=1111111110

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

5^{7} = 7^{5}

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

For a three-digit number where all three digits are the same, the number divided by the sum of the digits in the number is equal to 27.

To square a two-digit number with N in the tens place and 5 in the ones place, the answer is always (N^{2}+1) as the first two digits and 25 as the last two.

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