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Which primes multiply the following integrals? (three primes) 42,50,63
42=2*3*7
50=2*5*5
63=3*3*7
Quickly tell which prime numbers multiply the following numbers. 15.22.25.42.45.63
15=3X5
22 = 22
25=5X5
42=2X3X7
45=5X3X3
63=7X3X3
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
It is known that the left and right focal points of the hyperbola x square / 16-y square / 9 = 1 are F1 and F2 respectively. If a point on the hyperbola satisfies the absolute value of AF1 * the absolute value of af2 = 20,
According to the definition of hyperbola
There are: | AF1 | - | af2 | = 2 √ 16 = 8
And: | AF1 | * | af2 | = 20
When | AF1 | > | af2 |
After sorting out the above two formulas, we get that | AF1 | - | af2 | = 8 | AF1 | * | af2 | = 20
The solution of the equations is | AF1 | = 10 | af2 | = 2
When MF1|
If the left and right focal points of hyperbola x ^ 2-4y ^ 2 = 4 are F1 and F2 respectively, and the right branch of the straight line crossing through F2 is at a and B, if | ab | = 5, then the perimeter of △ af1b is?
There is something wrong with the one upstairs
Standard form X ^ 2 / 4 - y ^ 2 = 1, a = 2, 2A = 4
therefore
F1A-F2A=4
F1B-F2B=4
Add
F1A+F1B-AF2-F2B=F1A+F1B-AB=8
So F1A + F1B = 13, perimeter = 13 + 5 = 18
If the focus of hyperbola & nbsp; x2-4y2 = 4 is F1, the left branch of the intersection of F2 through F1 is a and B, if | ab | = 5, then the perimeter of △ af2b is______ ．
According to the meaning of the title, | af2 | - | AF1 | = 2A = 4 & nbsp; & nbsp; ① | BF2 - | BF1 | = 2A = 4 & nbsp; & nbsp; ② and | ab | = 5, ① + ② get: | af2 | + | BF2 | = 13, perimeter is 18, so the answer is: 18
Let F1 and F2 be the two focal points of the hyperbola x2-4y2 + 16 = 0, and the straight line passing through point F2 intersects the hyperbola at two points a and B, then AF1 + BF
Then AF1 + bf1-ab =? It's best to match the pictures. The process must be effective
How much more BF?
The hyperbola is y ^ 2 / 4-x ^ 2 / 16 = 1
a^2=4 a=2 b^2=16 b=4
Suppose a and F1 are above the y-axis, B and F2 are below the y-axis
AF2-AF1=2a=4
BF1-BF2=2a=4
AF1=AF2-4
BF1=BF2+4
AF1+BF1-AB=2BF2
It is known that the center of the ellipse is at the origin, the two focal points F1 and F2 are on the x-axis and pass through the point a (- 4,3). If F1A ⊥ F2a, the standard equation of the ellipse is obtained
Let the standard equation of ellipse be x2a2 + y2b2 = 1 (a > b > 0); ∵ F1A ⊥ F2a, ∵ F1A · F2a = 0, ∵ (- 4 + C, 3) · (- 4-C, 3) = 0, which is reduced to 16-c2 + 9 = 0, the solution is C = 5. Simultaneous 16a2 + 9b2 = 1A2 = B2 + 52, the solution is A2 = 40b2 = 15. Therefore, the standard equation of ellipse is x240 + y215 = 1
Point P is on the hyperbola e e: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 with F1 and F2 as the focus. It is known that Pf1 is perpendicular to PF2, the module of Pf1 is equal to the module of double PF2, and O is the coordinate origin. (1) eccentricity of spherical hyperbola e e (2) straight lines passing through point P intersect at P1 and P2 respectively, and vector OP1 multiplies vector op2 = - 27 / 4, double vector PP1 + vector PP2 = zero vector, For the equation (3) of hyperbola e, if the line L passing through Q (m, 0) intersects two points m and N of hyperbola E in the second question, and the vector MQ = α vector QN, ask whether there is a fixed point G on the X axis, so that the vector F1F2 is perpendicular to (vector GM - α vector GN)?
1、∵PF1-PF2=2a ,PF1=2PF2
∴PF2=2a,PF1=4a
Brought in: PF2 ^ 2 + Pf1 ^ 2 = 4C ^ 2
B = 2A
e=c/a=(a^2+b^2)^0.5/a=√5
The curve y = | X & # 178; - 4x + 3 | has two different intersections with the straight line y = M. how to find the value range of the real number m?
The key to solve the problem is to draw a curve,
Then we can draw the intersection point of X and y for x = Ӝ - 178; Ţ - 3
And, 1=
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