How many are the maximum four digits and the minimum four digits on the counter with five beads?

How many are the maximum four digits and the minimum four digits on the counter with five beads?

Maximum 9000, minimum 1004!

Can you use 6 beads to dial four digits on the counter? The biggest is () and the smallest is ()

6000 6

A three digit number is 0, and the number of hundred digits is less than that of ten digits. It takes 11 beads to dial this number on the counter. What is the three digit number?

Baiwei used (11 - 5) △ 2 = 3 pills
Ten used (11 + 5) △ 2 or = 11 - 3 = 8
So this three digit number is 380

Define the operation ⊙ a ⊙ B = AB + 2A + B on R, then the value range of real number x satisfying x ⊙ (X-2) < 0 is ()
A. {x | 0 ＜ x ＜ 2} B. {x | - 2 ＜ x ＜ 1} C. {x | - 2, or X ＞ 1} D. {x | - 1 ＜ x ＜ 2}

From the definition operation ⊙ a ⊙ B = AB + 2A + B, it is obtained that x ⊙ (X-2) = x (X-2) + 2x + X-2 = x2 + X-2.. x ⊙ (X-2) ＜ 0 ⇔ x2 + X-2 ＜ 0, the solution is: - 2 ＜ x ＜ 1. The value range of real number x satisfying x ⊙ (X-2) ＜ 0 is {x | - 2 ＜ x ＜ 1}

Define the operation ⊙ a ⊙ B = AB + 2A + B on R, then the value range of real number x satisfying x ⊙ (X-2) < 0 is ()
A. {x | 0 ＜ x ＜ 2} B. {x | - 2 ＜ x ＜ 1} C. {x | - 2, or X ＞ 1} D. {x | - 1 ＜ x ＜ 2}

From the definition operation ⊙ a ⊙ B = AB + 2A + B, it is obtained that x ⊙ (X-2) = x (X-2) + 2x + X-2 = x2 + X-2.. x ⊙ (X-2) ＜ 0 ⇔ x2 + X-2 ＜ 0, the solution is: - 2 ＜ x ＜ 1. The value range of real number x satisfying x ⊙ (X-2) ＜ 0 is {x | - 2 ＜ x ＜ 1}

If the power 0 of (x-1) - the power 0 of 2 (X-2) is significant, then the value range of X is () a.x > 1 b.x > 2 c.x ≠ 1 or X ≠ 2 D.X ≠ and X ≠ 2

D
The power 0 of (x-1) - the power 0 of 2 (X-2)
. x ≠ 1 and X ≠ 2

If the operation ⊙ a ⊙ B = AB + 2A + B is defined on R, then Xe (X-2) is satisfied

Scope: 1
The center point of a circle is B

If the cube of the polynomial ax of the letter x minus the square of 2x plus 6 plus the cube of (A-1) x plus 2bx minus 7 does not contain the first and third terms of X, the value of a and B can be obtained

ax^3-2x^2+6+(a-1)x^3+2bx-7
=(2a-1)x^3-2x^2+2bx-1
Without cubic term
Then 2a-1 = 0
a=1/2
It does not contain a linear term, that is, B = 0

A man uses a card with the letter "B", a card with the letter "a", two cards with the letter "g", three cards with the letter "e"
There are several wrong ways to spell "baggage"
Answer: 419
Thanks for your training

1. There are 7 kinds of arrangement for 7 cards
2. The letters A and g have 2! And 3! Repetitions, which need to be removed
3. Therefore, there are a total of 7! / (2! X3!) = 420 permutations, of which only one is correct, so there are many wrong spellings
420-1 = 419 species

Write the 11 letters "probability" on the 11 cards, and select 7 of them to calculate the probability that the result is ability
Wrong. It's probability

This is a lottery like thing. Let me talk about it. Probability actually has nine characters, including two B and one I, and one for the rest. If you specify any character to extract the probability that B and I are 2 / 11, the rest is 1 / 11. Because of the position relationship, it's like sports lottery, unlike welfare lottery string AB