There is an egg and an apple on the table

There is an egg and an apple on the table

there are an egg and an apple on the table.

The apple on the table is for you

The apple on the table is for you
The apple on the table is for you

Are there any apples on the table

There are some apples on the table

Test whether - 3 / 5,5 is the solution of equation 6x + 3 = 9x-12

Yes
.

How to calculate a mathematical problem
A late businessman left a will to his pregnant wife: if you give birth to a boy, give him two-thirds of the will, and you take one-third; if you give birth to a daughter, give her one-third of the will, and you leave two-thirds. But when a wife gives birth to a pair of twins, how should she share the estate?

If we look at this problem from a legal point of view, the will of the rich man is invalid. According to the legal succession order, the wife is 1 / 3, the son is 1 / 3, and the daughter is 1 / 3. If it's a question of intelligence, it's a question of proportion. The distribution ratio between the boy and the wife is 2:1, and the distribution ratio between the wife and the girl is 3:2

It is known that the first term A1 of the sequence {an} is 5, the sum of the first n terms is Sn, and S (n + 1) = 2Sn + N + 5

Let BN = an + 1
S(n+1)=2Sn+n+5——1
Sn=2S(n-1)+n-1+5=2S(n-1)+n+4——2
1-2
S(n+1)-Sn=2[Sn-S(n-1)]+1
a(n+1)=2an+1
a(n+1)+1=2(an+1)
b(n+1)=2bn
b(n+1)/bn=2
So {BN} is an equal ratio sequence of 2 with the first term of 6

6.5 * 2.1-0.65 and 3.12 + 31.2 * 9.9 have simple algorithm, how to calculate

6.5*2.1-0.65
=6.5*2.1-6.5*0.1
=6.5*（2.1-0.1)
=6.5*2
=13
3.12+31.2*9.9
=3.12+3.12*99
=312*(1+99)
=3.12*100
=312

If A1 = 1, A2 = 1-2, A3 = 1-2 + 3, A5 = A6 = A7 = a2007 = A2008=

a5=3 a6=-3 a7=4 a2007=1004 a2008=-1004
When n is odd, an = (n + 1) / 2
When n is even, an = - N / 2

Make up the smallest number with four ones
Use 4 1s to form a minimum number, which requires no addition, subtraction, multiplication and division. What is the minimum number? What is the maximum number?
Supplement: this problem lies in the power side

Maximum: the 11th power of 11, minimum: the 1st power of 1, the 1st power of 1

Let a and B be invertible matrices of order m and N respectively. It is proved that the block matrix [O A / b o] is invertible and the inverse is obtained

Reversible by a, B
Order H=
0 B^-1
A^-1 0
From H [o a; b o] = e
So [O A / b o] is reversible and [O A / b o] ^ - 1 = H