There is a math problem in the third year of junior high school, and a math problem in the third yea

Combine the two formulas and get it.

k/x=-x-6

x^2+6x+k=0

b^2-4ac=6*6-4k>0

So k<9 (and k is not equal to 0).

different intersections, so, b 2-4ac>0

Simultaneous equation, k x=-x-6

x^2+6x+k=0

b 2-4ac>0, i.e., 6*6-4*1*k>0, yields, k<0, and k is not equal to 0

Answer: k < 0, and k is not equal to 0

Column equation ha. Let k x=-x-6 make two different and non-zero solutions.

i.e. x2 6x k 0

So the square of 6 4k should be greater than zero.

So when k is less than 9 and x is equal to zero, k is equal to zero.

So k is not equal to zero.

So k is less than 9 and not equal to zero.

I'm sorry, but I couldn't see the minus ......sign before x in y=-x-6 in y=-x-6 last timeSimultaneous function, x 2+6x+k=0 (x<>0) has two different intersections, then the equation has two unequal roots, diaota>0 gives k<9

Or if you look at the image, when k<0 there is always a binary foci.

When k>0, there is only one intersection point when the hyperbola is tangent to y=-x-6, and the tangent point is (3,3).

3*3=9 x*y=k=9, there must be k<9 and k<>0

Solution: Knowing that PA is vertical OA, PB is vertical OB, OA=ob=4cm, and when OP is connected, the angle PAO=angle PBO=90°, and AOP and BOP have OP edge coincidence equally, so AOP is equal to BOP

So the angle opa = angle opb = 120° 2 = 60° so PA = 4 * root number 3 3 = 4 root number 3 3 (cm).

1、y=(240-2x)(x-50)=-2x2+340x-12000=-2(x-85)2+2450

2. When x=85, y has a maximum value of 2450

-2(x-85)

x-85)2=100→x-85=±10→

x = 95 or 75

Because the sales unit price shall not be higher than 90 yuan kg, so x=75 that is, the sales unit price should be set at 75 yuan.

4 root number 3

3cm four-thirds of the number three).

Link OA, OB

then oap= obp=90°

The sum of the inner angles of the quadrilateral is 360°

and apb=120°

aob=60°

poa=∠pob=30°

pa=4 root number 3=4 root number 3

3cm four-thirds of the number three).

If you don't write the topic clearly, what will others do?

The title should be:

The straight line Pa, Pb are the two tangents of the circle O, A, B are the tangent points, and apb=120°, the radius of the circle O is 4cm, find the length of the tangent Pa.

Let's start with two formulas:

Profit = Selling Price - Cost.

Profit Margin = Profit Cost.

Solution: The original cost is X, and Y technical innovation is carried out.

Original profit = 20% * x =

Original selling price = 20% *x + x =

So the selling price is the cost = x = times.

Current cost = x+(1%x * y) =x(1+y%)The current selling price remains unchanged.

Profit margin =

Profit margins fell by (12 11 % y) percentage points.

So the original profit margin 20% -**% point (12 11 % y) = current profit rate (

i.e. 20%-(12 11 % y) =

20%-(12/11 %y) = 1+y%)=> -12/11 %y)(1+y%)=> -12/11(1+y%)%y = y%=> 1

y = 10

So the company made 10 technological innovations.

Solution: The original cost is X, and Y technical innovation is carried out.

20%-(12/11 % y) =

20%-(12/11 %y) = 1+y%)=> -12/11 %y)(1+y%)=> -12/11(1+y%)%y = y%=> 1

y = 10

This question is usually the finale, the last question, it's very complicated, the first question seems to be equal to a little wait.

I can't help you anymore because I'm only a sophomore in junior high school.

You'll know the answer when you're done.

Solution: (1) The line y=-x+3 intersects the x-axis at the point b, when y=0, x=3, and the coordinates of the point b are (3,0).

And because the parabola passes through the two points A and B on the x-axis, and the axis of symmetry is x=2, according to the symmetry of the parabola, the coordinates of point A are (1, 0).

2) Because y=-x+3 passes through the point c, it is easy to know c(0,3), so c=3

And because the parabola y=ax 2+bx+c passes through the points a(1,0),b(3,0), then a+b+3=0,9a+3b+3=0

The solution yields a=1, b=-4

y=x^2-4x+3．

3) Connect Pb, by y=x 2-4x+3=(x-2) 2-1, get p(2,-1), let the symmetry axis of the parabola intersect the x-axis at the point m, in the right-angled triangle pbm, pm=mb=1, so pbm=45, pb=2

From the points b(3,0), c(0,3) ob=oc=3, in the isosceles right triangle obc, abc=45, from the Pythagorean theorem, BC=3 2

Suppose there is a point q on the x-axis, such that a triangle with points p, b, q as vertices is similar to the triangle abc

When bq bc=pb ab, pbq= abc=45, the triangle pbq triangle abc

That is, bq 3 2 = 2 2, bq = 3, and bo=3, the point o coincides with the point q, and the coordinates of q are (0, 0).

When qb ab=pb bc, qbp= abc=45, the triangle qbp triangle abc

i.e. qb 2= 2 3 2, so qb = 2 3

Because ob=3, oq=ob-qb=3-2 3=7 3.

So the coordinates of q are (7 3,0) 11 minutes).

So point q can't be on the x-axis to the right of point b

In summary, there are two points q (0,0) and q (7 3,0) on the x-axis, which can make the triangle with the vertices of points p, b, and q similar to the triangle abc

Solution: b(3,0) c(0,3).

There is an axis of symmetry that is x=2 to get -b a=2, and then bring in b c two-point coordinates a=-1 b=2 c=3

Bringing in the original function, when y=0, find two x coordinates that are different from b and a2It is known that 3 points can be found in the parabola.

Let the p point be (2,m), and on the image of y=x 2-4x+3, then:

m=2^2-4*2+3,m=-1.then PB= 2, CBO= PBO=45°.

From the Pythagorean theorem, it can be obtained: cb=3 2.

A triangle with vertices p, b, and q is similar to abc pb bq=ab bc or pb bq=bc ab, i.e., 2 bq=2 (3 2) or 2 bq=(3 2) 2 can be obtained: bq=3 or 2 3.

So there are two such points q on the x-axis:

1) When bq=3, q is at the origin, and the coordinates are (0,0);

2) When bq=2 3, oq=3-2 3=7 3, that is, the point q is (7 3,0).

The straight line y=-x+3 intersects the x-axis, the y-axis intersects at the point b, and the point c so b=(3,0).

c=(0,3)

And because the axis of symmetry is a straight line x=2

So y=x -4x+3

a=(1,0)

b=(3,0)

p=(2，-1)

q=(7, 3,0), p,b,q is a triangle with vertices, similar to abc.