Hurry ah Senior 1 Mathematics Compulsory 2 Exercises to Analyze Analyze Analyze Ah!!

The equation for a circle that passes through the point a(2,-1) and is tangent to the line x+y=1 and the center of the circle is on the line y=-2x.

Suppose the equation for the circle is (x-a) 2+(y-b) 2=r 2, and since the center of the circle is on y=-2x, so b=-2a, the equation for the circle is.

x-a)^2+(y+2a)^2=r^2.Since the circle passes through a(2,-1), there is.

2-a)^2+(-1+2a)^2=r^2………Equation 1 is tangent to the straight line x+y=1.

So there is: a+(-2a)-1|/√2=r………Equation 2 combines Equation 1 and 2 to solve a r to get the equation.

Solving the above equation yields: a=1, r=2

So the equation for the circle is :

x-1)^2+(y+2)^2=2

Let the center of the circle (a, -2a).

Then the circular equation (x-a) +y+2a) = r

Substitute a(2,-1) into the equation and get (a-2) +2a-1) =r =(a+1) 2 according to the formula for the distance from a point to a straight line

The solution of a=1 is that the circle equation is (x-1) +y+2) =2

Let the center of the circle be (x,-2x) The distance from the center of the circle to the straight line and the distance from point A are equal, list the equations, find x,y and write the equation according to the center of the circle and the radius.

Square p and q separately, remove ab and cd from the results, and the remaining part can be solved using the basic inequality.

Final answer: q>=p

From the Cauchy inequality, q= [mA+nc)*(b m+d n)] ma*b m)+ nc*d n)]2= (ab)+ cd)=p

i.e. q p (2 for squared).

Set to invest 10,000 in A and B in 10,000. Rule.

a>=2/3b

a>=25, b>=25, so 25<=b<=180a+b=300

Profit = < = 120+

So the maximum profit is 156

a+b=300 =>b=300-a

a>=b*2 3 =>a>=(300-a)*2 3 =>a>=120a>=25, b>=25 =>300-a>=25 =>a<=275 maximum, a=120, profit: 156

I don't know if it's right, refer to it.

Let the distance between the first buoy and the second buoy be A1, and the distance between the twelfth buoy and the thirteenth buoy is A12, then according to the title, it can be known:

a1+a12)*12]/2=1000

i.e. a1 + a12 = 500 3

According to the nature of the difference series: the sum of terms with the same distance from the first and last two terms is equal, i.e., a1 + a12 = a2 + a11 = a3 + a10 = ......

So the distance between the second buoy and the twelfth buoy is (500 3) * 5 = 2500 3

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Group A: 3 (1) (3) to (2) (4) (5) False 4 (1) 2) 8 (3) 2 (4) Parallel or in plane (5) Parallel or intersecting (6) Intersecting or different planes.

5 coplanar 6 because AA'//bb' ,aa'=bb'The quadrilateral aa'bb'is a parallelogram, so ab = a'b'In the same way, bc = b'c'So, 7 3; 3

Group B 1 (1)C (2)D (3)C

All of them are available.

I found it for you, the landlord.

But that's a non-txt file.,Can't be copied.,How to copy all kinds of letter formulas for you.,I'll see if it's OK.。

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Group p513(1) True (2) False (3) True (4) False (5) False 4(1) (2) 8 (3) 2 (4) parallel or in plane (5) parallel or intersecting (6) intersecting or different planes.

5.Coplanar. Am I right?

It is known that the function f(x) = (k is constant, e= is the base of the natural logarithm), and the tangent of the curve y= f(x) at the point (1, f(1)) is parallel to the x-axis.

) to find the value of k;

) to find the monotonic interval of f(x);

Let g(x)=(x2+x), where is the derivative of f(x), prov: for any x 0,.

Let the ratio be q, then a2=a1*q, a3=a1*q 2, a6=a1*q 5

Then: 2a1+3a2=2a1+3a1*q=1,(a1*q 2) 2=9*(a1*q)*(a1*q 5).

Simplification: 2a1+3a1*q=1, q 4=9*q*q 5

Q 2 = 1 9 is obtained by Eq, because the terms of the proportional series an are positive, i.e., q>0, so Q = 1 3, and substituting into the formula gets: A1 = 1 3

So the general formula for an is an=a1*q (n-1)=(1 3)*(1 3) (n-1)=(1 3) n (n 1).

1/bn=-2/[n(n+1)]=-2[1/n-1/(n+1)]

sn=-2[1-1/2+1/2-1/3+..1/n-1/(n+1)]=-2[1-1/(n+1)]=-2n/(n+1)

Let the midpoint be (m,n), then the distance between the midpoint and the fixed point is the root number (m-2) 2+(n+6) 2 chord center distance is the root number m 2+n 2 The distance between the center of the circle and the fixed point is the root number 40, then according to the valley theorem, there are (m-2) 2+(n+6) 2 +m 2+n 2=40 2m 2-4m+2n 2+12n=0 simplified (m-1) 2+(n+3) 2=10 (the trajectory is also a garden, and the center of the circle is (1,-3).