The actual output is 10% less than the planned output, so () is regarded as unit 1

The actual output is 10% less than the planned output, so () is regarded as unit 1

Planned output
The actual output is 10% less than the planned output, which is regarded as unit 1

"10% overproduction" is to regard "planned output" as unit "1". The "quantity of overproduction" is equivalent to 10% of "planned output"
Right or wrong? Ask for advice and explain why

Wrong! The planned output is 100
Overproduction 10

Three rational number addition and subtraction calculation problems, as long as 1-2 parts (drag type)
-1+2-3+4-5+6-···-99+100
1-2-3+4+5-6-7+8+9-10-11+···+1997-1998-1999+2000

-1+2-3+4-5+6-···-99+100＝（－1＋2）＋（－3＋4）＋.＋（－99＋100）＝1＋1＋1.＋1＝501-2-3+4+5-6-7+8+9-10-11+···+1997-1998-1999+2000 ＝（1-2-3+4）＋（5-6-7+8）＋.＋（1997-1998-1999+2000）＝0＋0＋....

Why is the sum of distances from a point on an ellipse to two focal points always 2A?

Ellipse formula: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0)
Two focus (- A, 0) (a, 0)
Let (x, y) be a point on an ellipse
Radical [(x + a) ^ 2 + y ^ 2] + radical [(x-a) ^ 2 + y ^ 2] = the sum of the distances from the point on the ellipse to the two focal points. The definition is 2A. We can directly substitute it for verification
The square error is as follows:
（x+a)^2 + y^2 + (x-a)^2 + y^2 +
2 radical [(x ^ 2 - A ^ 2) ^ 2 + y ^ 4 + y ^ 2 × [(x + a) ^ 2 + (x-a) ^ 2]]
= 2x^2 + 2y^2 + 2a^2 +
2 radical [(x ^ 2 - A ^ 2) ^ 2 + y ^ 4 + y ^ 2 × [2x ^ 2 + 2A ^ 2]] = 4A ^ 2
The transfer items are:
2x^2 + 2y^2 - 2a^2 =
2 radical [(x ^ 2 - A ^ 2) ^ 2 + y ^ 4 + y ^ 2 × [2x ^ 2 + 2A ^ 2]]
Square on both sides:
4x^4 + 4y^4 + 4a^4 + 8x^2×y^2 - 8x^2×a^2 - 8y^2×a^2=
4x^4 - 8a^2×x^2 + 4a^4 + 4y^4 + 8y^2×x^2 + 8y^2×a^2
Obviously, the above formula holds, so the sum of distances is 2a from the focal radius: Pf1 = a + ex0 PF2 = a-ex0 Pf1 + PF2 = 2A

The prime numbers within 20 are: 2, 3, 5, 7, 11, 13, 17 and 19. How do you calculate them

There is no number that can be divisible except 1 and itself

Given the point P (2,0) and the function y = x ^ 2-4, two points a and B on the image. (1) if the inclination angles of the line PA and Pb complement each other, it is proved that the slope of the line AB is a fixed value
Given the point P (2,0) and the function y = x ^ 2-4, two points a and B on the image
(1) If the inclination angles of PA and Pb are complementary, it is proved that the slope of AB is a fixed value;
(2) If ab ⊥ PA, find the abscissa range of point B

(1) Let a and B coordinate be (a, a ^ - 4), (B, B ^ 2-4) Pa, Pb slope be K1, K2 respectively, so K1 = (a ^ 2-4) / (A-2) = a + 2, K2 = (b ^ 2-4) / (b-2) = B + 2, because the inclination angle of PA and Pb is complementary, so K1 + K2 = 0, so a + B = - 4kab = (b ^ 2-4-a ^ 2 + 4) / (B-A) = a + B = - 4 = fixed value (2) because ab ⊥ PA

If M. n is on the two branches of the inverse scale function image, please point out the shortest segment Mn and find out the value range of segment Mn length

When m and N are symmetric about the center of the origin and the distance from m to the origin is the minimum, Mn is the minimum;
When m and N tend to infinity and infinity respectively, their length is the largest, which is also infinity + ∞
Set point m (x, y), minimum Mn = 2om = 2 √ (x ^ 2 + y ^ 2)
=2√(x^2+(k/x)^2)
≥ 2 √ 2 * x * k / x = 2 √ 2K (k also has a root)
If and only if x = K / x, that is, x = ± √ K
That is, the range of Mn length (2 √ 2K, + ∞)

If we know that the distance between two points a and B on the number axis and the origin is changed sign 2 and 3, then AB =?

So, ab = 3 - radical 2 or AB = 3 + radical 2

Let the probability density of random variable (x, y) be f (x, y) = {K (6-x-y), 0

If ∫ f (x, y) DXDY = 1, k = 1 / 8 can be obtained
P {x + y ≤ 4} = ∫∫ f (x, y) DXDY = ∫ DX ∫ K (6-x-y) dy = 2 / 3, (the lower upper limit of the former integral is 0 and 2, and the lower upper limit of the latter integral is 2 and x-4)
The determination of integral limit needs to draw 0

Solving quadratic equation with one variable by MATLAB
For example, for the equation T + 9.8 * T ^ 2-2 = 0, how can the answer be expressed in the form of a root

1、y=solve('9.8*t^2+t-2')
2、y=solve('98*t^2+10*t-20=0')
3、a=[9.8 1 -2]; roots(a)
Try the difference between the first and the second!