After Mr. Li transferred one girl from the math interest group to the English interest group, 17 of the remaining students were girls. If two boys were transferred instead of the girl, 15 of the remaining students were girls. Q: how many students were there in the math interest group?

Given A2 + A-1 = 0, how much is A3 + 2A2 + 3? 2 and 3 after a are the sum of squares cube

From A2 + A-1 = 0, we can get A2 + a = 1. By sorting A3 + 2A2 + 3 into a (A2 + a) + A2 + 3, and substituting A2 + a = 1, we can get a (A2 + a) + A2 + 3 = A2 + A + 3 and A2 + a = 1, so A2 + A + 3 = 1 + 3 = 4

The straight line passing through the right focus F of hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) intersects the hyperbola at two points m and N, intersects the Y axis at P, and finds the value of PM / mf-pn / NF?

Thank you, but the answer is 2A ^ 2 / b ^ 2

Mathematical problems of Qin Jiushao's algorithm
Using Qin Jiushao's algorithm to find the value of polynomial f (x) = 4x ^ 5 + 3x ^ 4 + 2x ^ 3-x ^ 2-x-2 / 1 when x = - 2 is ()

Qin Jiushao algorithm is a polynomial simplification algorithm proposed by Qin Jiushao, a mathematician in the Southern Song Dynasty. It is called Horner algorithm or Horner scheme in the West and named after William George Horner, a British mathematician
Rewrite a polynomial f (x) = a [n] x ^ n + a [n-1] x ^ (n-1) +. + a [1] x + a [0] as follows:
f(x)=a[n]x^n+a[n-1]x^(n-1))+.+a[1]x+a[0]
=(a[n]x^(n-1)+a[n-1]x^(n-2)+.+a[1])x+a[0]
=((a[n]x^(n-2)+a[n-1]x^(n-3)+.+a[2])x+a[1])x+a[0]
=.
=(.((a[n]x+a[n-1])x+a[n-2])x+.+a[1])x+a[0].
When calculating the value of the polynomial, first calculate the value of the first degree polynomial in the innermost bracket, that is
v[1]=a[n]x+a[n-1]
Then, the value of the first-order polynomial is calculated layer by layer from the inside out
v[2]=v[1]x+a[n-2]
v[3]=v[2]x+a[n-3]
.
v[n]=v[n-1]x+a[0]
In this way, finding the value of polynomial f (x) of degree n is transformed into finding the value of polynomial of degree n
(Note: numbers in brackets indicate subscripts)
Conclusion: for a polynomial of degree n, n times multiplication and N times addition can be done at most
Substitution calculation:
v[1]=a[n]x+a[n-1]=4*（-2）+3=-5
v[2]=（-5）*（-2）+2=12
v[3]=12*（-2）-1=-25
v[4]=（-25）*（-2）-1=49
V [5] = 49 * (- 2) - 1 / 2 = - 98 and 1 / 2
Using Qin Jiushao's algorithm to find the polynomial f (x) = 4x ^ 5 + 3x ^ 4 + 2x ^ 3-x ^ 2-x-2 / 1, when x = - 2, the value is (- 98 and 1 / 2)

It is known that point O is the symmetry center of rectangle ABCD, the line L passing through point O intersects line ad at m, on intersects line DC at n perpendicular to OM, and connects with Mn. Now rotate line L clockwise around point O. (1) as shown in Figure 1, when point m n is on edge ad CD respectively, please write out the equivalent relationship between line am Mn CN: --; (2) as shown in Figure 2, when point m n is on the extension line of edge AD and CD respectively, Please write out the equivalent relationship between the segments am Mn CN and prove it

1) Mn ^ 2 = am ^ 2 + CN ^ 2 extends Mo, intersects BC with G and connects GN. Because o is the center of symmetry, CG = am, OM= OG.ON It's the vertical bisector of GM, so Mn= GN.AM ^2 + CN ^ 2 = CG ^ 2 + CN ^ 2 = GN ^ 2, so Mn ^ 2 = am ^ 2 + CN ^ 22) Mn ^ 2 = am ^ 2 + CN ^ 2 is still valid. It is proved that extending the CB extension line of Mo intersection to g, connecting GN

Sequence 3, 7, 13, 21, 31 The general formula of is ()
A. An = 4n-1b. An = n3-n2 + N + 2C. An = N2 + N + 1D

Let this sequence be {an}, then a2-a1 = 7-3 = 4, a3-a2 = 13-7 = 6, a4-a3 = 21-13 = 8, a5-a4 = 31-21 = 10 ，∴an-an-1=2n，∴an=（an-an-1）+（an-1-an-2）+… +（a2-a1）+a1=2n+2（n-1）+… +2 × 2 + 3 = 2 × n (1 + n) 2 + 1 = N2 + N + 1

If point P is in the same plane as line a and B, and PA = 5cm, can you determine the length of line Pb

No, absolutely not. The angle between PA and ab is unknown, so the length of Pb is uncertain

X-1 / 6 [36-12 (3x + 1 / 5)] = 1 / 3 X-2

X-1 / 6 [36-12 (3x + 1 / 5)] = 1 / 3 (X-2)
X - [6-2 (3x + 1 out of 5)] = 1 out of 3 (X-2)
X - [6x-2 of 6-5] = 1 of 3 (X-2)
X - [6x of 4-5] = 1 / 3 (X-2)
11x-4 / 5 = 1 / 3 (X-2)
3(11x-4)=5(x-2)
33x-12=5x-10
28x=2
x=1/14

The bottom radius of a cylindrical measuring cup for water is 5cm long. There is a cone with a bottom area of 30cm2 in the cup, which is completely in water. If you take out the cone, you can reduce the height of the cone by 2cm

After taking out the cone, the water level drops by 2cm,
It shows that the volume of the cone = the bottom area of the cylinder measuring cup * 2cm = π * 5 & # 178; * 2 = 50 π≈ 157 (CM & # 178;)
So 1 / 3 * 30 * H = 157
The solution is h = 15.7 (CM)

After Mr. Li transferred one girl from the math interest group to the English interest group, 17 of the remaining students were girls. If two boys were transferred instead of the girl, 15 of the remaining students were girls. Q: how many students were there in the math interest group?