If a is a natural number greater than 1, then the square of a must be () A. prime B. composite C. odd D. even

If a is a natural number greater than 1, then the square of a must be () A. prime B. composite C. odd D. even

B
If you understand and solve your problem,

N is a natural number greater than 1. What is the square of N? Even? Prime? Composite?

Total number

The sum of two prime numbers is 40. What is the maximum product of the two prime numbers?

Prime numbers less than 40 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, of which the sum of two prime numbers is 40 is 3 and 37; 11 and 29, 17 and 23, their products are: 3 × 37 = 111; 11 × 29 = 319; 17 × 23 = 391; a: the maximum product of these two prime numbers is: 391

Let the center of the ellipse C1 be at the origin, and its right focus coincides with the focus F of the parabola C2: y ^ 2 = 4x. A line passing through F and perpendicular to the X axis intersects with C at two points a and B, and intersects with C2 at two points c and D. It is known that | CD | / | ab | = 4 / 3
(1) The equation of finding ellipse C1
(2) If | PQ | / | Mn | = 5 / 3, the equation of line L is obtained
The first question has been made. The elliptic equation is (x ^ 2) / (4) + (y ^ 2) / (3) = 1
The ellipse a = 2, B = √ 3, C = 1, f coordinate is (1,0)
My idea is to set the equation y = K (x-1) of the straight line L, substitute it into the equation of ellipse and parabola to calculate the intersection point, and then substitute the distance formula into the ratio to calculate the slope. But it's a huge amount of calculation. Do you have any simple algorithm

The chord length formula = √ (1 + K ^ 2) ×√ [(x1 + x2) ^ 2-4x1x2] is used to calculate the distance between two points
Let the equation of straight line be: x = KY + 1, simultaneous y ^ 2 = 4x, eliminate the parameter x to get: y ^ 2-4ky-4 = 0, the discriminant is: 16K ^ 2 + 16 > 0, and then combined with the relationship between root and coefficient, there is: | PQ | = √ (1 + K ^ 2) × √ (16K ^ 2 + 16) = 4 (k ^ 2 + 1), similarly, simultaneous (x ^ 2) / (4) + (y ^ 2) / (3) = 1 has: (3K ^ 2 + 4) y ^ 2 + 6ky-9 = 0, the discriminant is 36K ^ 2 + 36 (3K ^ 2 + 4) > 0, |Mn | = [12 (k ^ 2 + 1)] / (3K ^ 2 + 4) and | PQ | / | Mn | = 5 / 3, that is: 4 (k ^ 2 + 1): [12 (k ^ 2 + 1)] / (3K ^ 2 + 4) = 5 / 3, the solution is: k = ± √ 3 / 3 and a straight line crosses the point F (1,0), then the equation is: y = √ 3x - √ 3 or y = - √ 3x + √ 3

If a divided by B equals 3, then a is a multiple of B. right or wrong
If it's not an integer, it should be a multiple, right?

No, 0.6 divided by 0.2 = 3, 0.6 is not a multiple of 0.3
The factors and multiples studied in primary school are all integers

It is known that the parabola y = x ^ 2-2x + m intersects the X axis at points a (x1,0), B (x2,0)
Given that the parabola y = x ^ 2-2x + m intersects with the X axis at a (x1,0), B (x2,0) (x1 & gt; x2), let the vertex of the parabola y = x ^ 2-2x + m be m. if △ AMB is a right triangle, find the value of M

Y = (x-1) ^ 2-1 + m m (1, m-1) X1 + x2 = 2 X1 * x2 = m △ AMB is a right triangle, only am ^ 2 + BM ^ 2 = AB ^ 2 (x1-1) ^ 2 + (m-1) ^ 2 + (x2-1) ^ 2 + (m-1) ^ 2 = (x1-x2) ^ 2 x1x2 - (x1 + x2) + 1 = (m-1) ^ 2 m ^ 2-3m + 2 = 0, M = 2 / 1, X1 does not equal to x2 M = 2

A number divided by 4 plus 3 equals the number divided by 5 plus 4. What is the number? A number divided by 4 plus 3 equals the number

Let this number be X
X/4+3=X/5+4
5X+60=4X+80
5X-4X=80-60
X=20
20/4+3=8

Why 3x ^ 2 + 11xy + 10Y ^ 2 = (3x + 5Y) (x + 2Y)

3x^2+11xy+10y^2
=3x ^ 2 + 6xy + 5xy + 10Y ^ 2 11xy is converted into two terms
=3x (x + 2Y) + 5Y (x + 2Y) grouping decomposition
=(3x+5y)(x+2y)

3＋7＋11＋15＋19＋…… How much is + 91 + 95 + 99
Help!
Why + 1
What about it

This is the summation problem of arithmetic sequence, the formula is: (first term + not term) * number of terms / 2
But we need to find the number of items first: number of items = (last first) / tolerance + 1
First find how many numbers to add, that is, the number of terms = (99-3) / 4 + 1 = 25
Sum again: (3 + 99) * 25 / 2 = 1275

The equations 2Y + 3x = 1,
It's not the one above, it's this 5x-2y-1 = 0, 2x-3y = 2

(1) 5x-2y-1 = 0 (after sorting: 5x-2y = 1; both sides × 3, get (3) (2) 2x-3y = 2 (both sides × 2, get (4) (3) 15x-6y = 3 (4) 4x-6y = 4 (subtract from (3), 11x = - 1x = - 1 / 11, substitute into (2) 2 * (- 1 / 11) - 3Y = 2-3y = 2 + 2 / 11y = 24 / 11 * (- 1 / 3) = - 8 / 11, check (1) 5 * (-)