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Choose a CA with at least one white ball: one white ball or two white balls, red:white = 1:1 or 0:2 are both white balls: there are two white balls, red:white = 1:1

It can be established at the same time, so it is not mutually exclusive.

b has at least one white ball: one white ball or two white balls, red:white = 1:

1 or 0:2 at least one red ball: one red ball or two red balls, red:

White = 1:1 or 2:0 can be held at the same time, so they are not mutually exclusive.

c has exactly one white ball: there is one white ball, red:white = 1:1 exactly two white balls: there are two white balls, red:white = 0:2

These two events cannot be held at the same time, so they are mutually exclusive.

d has at most one white ball: there are two cases: there is a white ball without a white ball, red:white = 1:1 or 2:0

Both are red balls: including a case: two red balls, i.e. no white balls, red:white = 2:0 so these two events can be held at the same time and are not mutually exclusive.

As for the opposing event, that is, if A and B are opposing events, then one of A and B must occur, and they cannot occur at the same time. If red:white=2:0 in option c, neither a nor b happens, so the two events in c are mutually exclusive and non-opposite.

The answer is D, at most it means the most, such as at most one white ball, that is, there is a white ball or none of the white balls. Complete.

Summary. Please describe your problem more specifically, you can use ** and other forms, and talk to the teacher in detail, so that the teacher can better help you.

Please describe your problem more specifically, you can use ** and other forms, and talk to the teacher in detail, so that the teacher can better help you.

How to do the second question? To break down an equation, don't recursively equation.

Dear, you can also list equations, I don't know if you need that method<><

These are two different methods, and you can do either of them <><

Thank you, can you give me two questions?

You can send us the question and read the <>

I'll help you figure out the third question, and you'll <>< the rest

No, you've already solved this problem.

The teacher asked me to tell him, do you think you can talk about it? I will give you a thumbs up, I will come back to you next time.

Okay, okay, I'll tell you about my thinking.

In the third question, let's take an example, 123 is equal to the hundred-digit macro wheel hunger 1, the ten-digit number is Tonglimb 2, and the single digit is 3, so 123 1 100+2 10+3 1

In the fourth question, the four shaded parts can be directly spliced together into a rectangular <>< that is less than 2 meters in length and width

Let xy equal to 0 respectively to find the intercepts are k 3 and k 4, add them to get 2, and solve k = 24 7.

Let the intercept on the x-axis be m, and the intercept on the y-axis will be (2-m), so 3m-k=0

4x(2-m)-k=0

Two equations are connected.

It is derived from m=8 7

So k = 24 7

When x is equal to 0, y is equal to a quarter of k

When y is equal to 0, x is equal to one-third of k

A quarter of a k plus a third of a k equals 2

k equals.

Substitute x=0 and y=0 respectively.

We get y=k4 and x=k3

Because the sum of the intercepts is 2

Hence k 4+k 3 = 2

The solution is k = 24 7

Using the fixed score point formula, p(2,-7) has a fixed ratio of -1 2 (because p1p and pp2 are in opposite directions) to substitute the formula, determine that it is right, and you can calculate it.

Let p2 be (x,y) p1 be (x1,y1) p be (x2,y2), then x2=-1=(-4-1 2y) (1-1 2)=2. The same is true for y2. Then just check it once.

If there is a mistake in the question, there should be four coordinates, otherwise it will not work.