Help solve some math problems and get a high score

1, the equation of the circle is x 2 + y 2-4x-4y-10 = 0r = 3 2, and the center of the circle is (2, 2).

The line is separated from the circle by a difference of 2r=6 2

2. The description is not clear, and there are many tangents that have not been said.

One thing that can be determined is the y-axis, which is x=0

3. What does "what is the value range of the straight line l"?

To satisfy the condition, the distance d from the center of the circle to the straight line is satisfied.

0 d 2 is acceptable.

4, chord ab = 2 3

then the distance from the center of the circle to the straight line is 1

2a-2+3|/√（a＾2+1）=1

a=0 or a=-4 3

5. "Circles with radius 1 and positive semi-axis respectively" which axis is the positive semi-axis.

The positive semi-axis of the x-axis: The equation for the circle is (x-2-3) 2+(y-1) 2=1y-axis: the equation is (x-1) 2+(y- 3) 2=16, d=|-15+5|/√10＞r

0＜r＜√10

7，5d=30-15，d=3

8，a1+a2+..an=2＾n-1

9,10 If you don't know the topic, please add.

1.The center of the circle (, the distance to the straight line is 5 2, the radius of the circle is 3 2 is less than 5 2, so the maximum distance is 2 2, the minimum distance is 8 2, and the difference is 6 2, which is equivalent to 1 diameter.

2.There are a lot of tangents, you didn't miss the conditions?

3.That is, the distance between the line and the center of the circle is within 2, so a 2+4ab+b 2 0

It's not a straight line.

5.If tangent to the y-axis, it is (x-1) 2+(y- 3) 2=0, and if tangent to the x-axis, it is (x- 3 2) 2+(y-1) 2=06i.e. (-5,0) is greater than the radius from the straight line, so 0 r 10d = 3

8.Proportional series Answer=2 n-1

9.What are you talking about, this is the meaning of the topic.

10.Original = 5 + y x + 4x y 5 + 4 = 9

Are you lazy and don't want to do your homework?

Now there are n people, m people come every minute, and each gate per minute can let s people enter the park (n+45m) 3s=45 ==> n=135s-45m(n+20m) 12s=20 ==> n=240s-20m now requires (n+xm) 8s=x ==> x in 8s=n x+m, x=n (8s-m).

a(135s-45m)+b(240s-20m)=c(8s-m) ==> 135a+240b=8*(45a+20b) ==>225a=80b

One set of possible solutions is: a=16, b=45

So 61n 12960s 1620m so x=1620 61=This is the answer (although the number is a bit strange, but if the number is correct, it should be this) In addition, is this park a park?? It takes so long to enter the park and there are so many people going...

Set Y people entering every minute, the original X people, and A people leaving each door every minute, X+45Y=3*45*A

x+20y=12*20*a

The solution is: x=324a, y=-21a 5

So the time it takes to complete the 8 doors can be calculated:

x+ty=8*t*a

324a-21at/5=8at

324-21t/5=8t

t = 1620 61 = minutes = 26 minutes 33 seconds.

It takes 26 minutes and 33 seconds to open 8 doors.

There are s people in the park, x tourists enter every minute, and each gate can walk y people every minute. So:

This question is not quite correct, it should be that 12 doors are reduced by 1 4 minutes at least compared to 3 gates, i.e. less than 45 4 minutes. And 20 minutes is not reasonable.

If it is 6 doors for 20 minutes, then.

s = 45 * 3y-x ) Equation 1s = 20* (6y-x) Equation 2 Solution Equation 1, Equation 2Get: x= , s = 108y

Open 8 doors, then s = t * 8y-x ) then it takes time t to be 108 minutes.

If each gate can have x people out per minute, and the fixed number of tourists entering per minute is y, then 3·45·x + 45·y = 12·20·x + 20y, that is, (240-135)x = (45-20)y21x=5y

x=5/21

It takes t minutes to open 8 doors, then.

3·45·x + 45·y = 8·t·x + t·yt = (140x + 45y)/(8x+y)= (140·5/21 + 45 )/(8·5/21+1)= (700+45·21)/61

27 (min).

Suppose the original x people, n people come every minute, and a door passes k people every minute, then 45*k*3=x+45*n and 12*k*20=x+20*n, so we can know n=-21*k 5

Then 8 people set the time to t, then t*k*8=x+t*n, bring in the calculation to know that t=1620 61 is about 15 minutes is impossible, 12 doors take 20 minutes.

There is a problem with this question, if you change 20 minutes to 10 minutes, you should be able to do it.

From the sinusoidal theorem, the angle c = 90 degrees can be obtained.

So the angle a = 30 degrees.

Then use the sine theorem.

You can make a tease to get a=1

Or directly from the 30-degree angle to the side of the side of the scattered Hu rush to the right-angle side of the one and a half of the annihilation.

Sine theorem. a/sina=b/sinb=c/sinc

2+2^2+2^3+2^4+..2^n=2^(n+1)-2=1000 000

n=log2(999998)-1=

To give you an estimate, 2 10 = 1024, 2 20 = 1024 10 6 so n + 1 20, n 19

Let f(x)=ax 2+2x+1

f(0)=1＞0

There must also be a negative root, so the parabola must be open downward, i.e. a 0 and then as long as the parabola has an intersection point with the x-axis.

0 and a 0 solves: a 0

The next question can ignore the distance from (-1,2) to y=1 2x.

d=√5

1. When a>0, 0 and when x=0, the function y<0, find a. No solution.

2. When a=0, the function is 2x+1=0, which is in line with the topic.

3. When a<0, 0 and when x=0 function y>0, find a, this has a solution.

The value of a is the union of 2 and 3.

On the phone, you can't afford to play, you can only write the steps, you just perfect it.

This is the same as the textbook examples.

The distance from point p to point f(, which is 1 greater than the distance of his straight line x=-2, means that the distance from point p to point f( is equal to its distance from x=-3, then it satisfies the parabolic definition.

The opening is to the right, p 2 = 3, so 2p = 12The equation is as follows: y 2 = 12x