Boys account for 5 / 9 of the class, while girls account for ()% of the boys, and girls are ()% less than boys The formula also needs to be improved Equations don't need to be equations, just formulas

Boys account for 5 / 9 of the class, while girls account for ()% of the boys, and girls are ()% less than boys The formula also needs to be improved Equations don't need to be equations, just formulas

Boys account for 5 / 9, and the ratio of boys to girls is 5:4
Girls account for 4 / 5 of boys = 80%
Less students than boys (5-4 /) / 5 = 20%

The number of boys in class 61 is two-thirds of that of girls. What percentage of girls in the class?

The proportion of female students in the whole class = 1 ÷ (1 + 2 / 3) = 60%

The number of boys in class 61 is 7 / 9 of the number of girls, and the number of girls accounts for ()% of the class

If the number of boys is x and the number of girls is y, then the total number is x + y. then the number of boys is x = 7 / 9y, and the proportion of girls in the class is Y / x + y = Y / (7 / 9y) + y = 9 / 16 = 56.26%

Calculation method and answer of 3 / 8 - (2 / 5-5 / 8)

3/8+5/8-2/5=3/5

In trapezoidal ABCD, ad ‖ BC, ab = CD, diagonal BD intersects median EF at g, eg ∶ GF = 1 ∶ 3, ad = 2, ∠ DBC = 45 °, then the area of trapezoidal ABCD is =?

∵ eg = 1 / 2ad = 1 / 2 × 2 = 1 ∵ eg: GF = 1:3 ∵ GF = 3eg = 3 × 1 = 3 ∵ BC = 2gf = 6 through D as DH ⊥ BC in H ∵ DBC = 45 °, DH = BH = bc-ch = BC - [(BC-AD) / 2] = 6 - [(6-2) / 2] = 4, so the area of trapezoidal ABCD is equal to: (AD + BC) × DH

6 × 6 = 1 () 6 = 6 ÷ () 8, fill in the appropriate number in the box to make the equation hold

6×6=1（8） 6=6÷（1）8
This question should be upside down. Look at it upside down, 81 = 9 × 9,81 △ 9 = 9

Let a be a square matrix of order n over the number field F with rank a = 1. It is proved that (1) there exist n * 1 matrix and 1 * n matrix C such that a = BC (2) a ^ 2 = Ka

1. R (a) = 1, there are invertible n-order square matrices P and Q, a = pe11q, E11 is the matrix with the first row and first column elements = 1 and other elements = 0. A = P (1,0,..., 0) ^ t (1,0,..., 0) Q
B=P（1,0,...,0）^T,C=（1,0,...,0）Q
A=BC
2. CB = (k) in fact, K is the sum of the products of the first row of Q and the corresponding elements of the first column of P
A^2=BCBC=B(K)C=KBC=KA

There is a two digit ten digit number a, and the one digit number is 2 larger than the ten digit number. What is the two digit number?

The single digit is a + 2
So this two digit number is 10A + A + 2, that is 11a + 2

Let the three eigenvalues of the third order square matrix a be: λ 1 = 2, λ 2 = - 1, and λ 3 = 3, then the determinant | a * | corresponding to the adjoint matrix of a is__________
Let the three eigenvalues of the third order square matrix a be: λ 1 = 2, λ 2 = - 1, and λ 3 = 3, then the determinant | a * | corresponding to the adjoint matrix of a is______________ .

Because a * a ^ * = | a | e is a determinant on both sides
|A | * | a ^ * | = | a | ^ 3 (the upper corner is marked with 3, because it is a matrix of order 3)
|A^*|=|A|^2
The determinant of matrix A is the product of eigenvalues, that is | a | = 2 * (- 1) * 3 = - 6
So | a ^ * | = (- 6) ^ 2 = 36

Given that two circles x2 + y2 = 10 and (x-1) 2 + (Y-3) 2 = 20 intersect at two points a and B, then the equation of line AB is______ ．

Because two circles intersect at two points a and B, the coordinates of two points a and B satisfy both the equation of the first circle and the equation of the second circle. By making a difference between the two circle equations, the equation of straight line AB is: x + 3Y = 0, so the answer is & nbsp; X + 3Y = 0