Simple junior high school math problems, simple junior high school math problems

The answer is 2 3

The solution is: the square of ac plus the square of bc is equal to the square of ab (this is the formula) bc is one-half of ab. (Right triangle, the opposite side of 30 degrees is one-half of the hypotenuse) so the square of ac is equal to the square of ab minus one-half of the square of ab and the square of ab is equal to 4, so ab is equal to 2 3

AB is hypotenuse!!

The first type: solution: angle a = 30 degrees cos30 ° = 3 2 = ac abac=2 ab = 4 3 3 (4/3 * root number 3) The second type:

Solution: c=90° a=30° that b=60°

sin b = 60° = hypotenuse = ac ab 3 2 ditto: get ab = 4 3 3

And so on......

In the triangle abc, the angle c = 90 degrees angle a = 30 degrees, so cos angle a = ac ab = 2 ab, so ab = 4 times the root number 3 3

In RT ABC, ab root number ac bc 4 ab ab four-thirds root number 3 The side opposite at the 30° angle is equal to half of the hypotenuse. Use the Pythagorean theorem to find ab

1、y=30-2x (0=x(30-2x)

2(x²-15x+

When x = m, s has a maximum value of square meters.

3. S 88 is available:

x(30-2x)≥88

x-11)(x-4)≤0

Solution: 4 x 11

And there is y 18, i.e., 30-2x 18 solution: x 6, and the sum up: 6 x 11

Solution: (1) y=30-2x(6x<15).

2) Set the area of the rectangular nursery garden as s

then s=xy=x(30-2x)=-2x2+30x s=-2(

Knowing from (1), 6 x < 15

When x=, s is the maximum

That is, when the rectangular nursery garden is perpendicular to the side of the wall with a length of meters, the area of this nursery garden is the largest, and the maximum value is (3)6 x 11

1：y+2x=30（0，15）

2: s=xy=x*(30-2x)=-2x squared + 30x, when x=-b 2a=15 2, s is 225 2 maximum

3: Do the math, bring s=88 in...

Solution: Problematic x 2-2x=5 => x 2-2x+1=6 =>(x-1) 2=6 =>x=1+root number 6 or x=1-root number 6

x 2 - root number 3x-1 = 0 = > x 2 - root number 3 + 3 4 = 7 4 = > (x - root number 3 2) 2 = 7 4

x=(root3-root7) 2 or x=(root3+root7) 2

y^4+y^2-20=0 => (y^2+5)(y^2-4)=(y^2+5)(y-2)(y+2)=0 y^2+5 =/= 0

y=2 or y=-2

3-k)(2-k)x^2-(24-9k)x+18=0 => (3-k)x-3)((2-k)x-6)=0

x=3/3-k x=6/2-k

So k=0 k=4

When k=2 k=3 x has only one root, so take.

Finally, k=0 k=4 is solved

3.Because the equation for the unary quadratic equation (3-k)(2-k) x 2-(24-9k)x+18=0 about x has two roots: so.

(24-9k) 2-72(3-k) (2-k) greater than or equal to 0

Solve the range of k: again because both are integers.

Leave the part of the root less than or equal to 0, and keep the part greater than or equal to 0 is what you want!! With utid, you are calculating to deepen your memory!! I've been writing for so long!! It's late at night, may you have a good dream!! Learning Progress!!

1 x 2-2x=5 => x 2-2x+1=6 =>(x-1) 2=6 =>x=1+root number 6 or x=1-root number 6

x 2 - root number 3x-1 = 0 = > x 2 - root number 3 + 3 4 = 7 4 = > (x - root number 3 2) 2 = 7 4

x=(root3-root7) 2 or x=(root3+root7) 2

y^4+y^2-20=0 => (y^2+5)(y^2-4)=(y^2+5)(y-2)(y+2)=0 y^2+5 =/= 0

y=2 or y=-2

2 3 (3-k)(2-k)x^2-(24-9k)x+18=0 => (3-k)x-3)((2-k)x-6)=0 =>

x=3 3-k x=6 2-k so k=0 k=4 when k=2 k=3 x has only one root, so you can't want it.

Finally, k=0 k=4 is solved

1. x 2-2x=5, (x-1-root number 6) (x-1 + root number 6) = 0. So: x1 = 1 - root number 6, x2 = 1 + root number 6

x 2 - (root number 3) x - 1 = 0; [x-(root3+root7) 2] [x-(root3+root7) 2]=0so:x1=(root3+root7) 2;x2=(root3-root7) 2

y^4+y^2-20=0 (y^2-4)(y^2+5)=0 (y-2)(y+2)(y^2+5)=0

so y=2 or y=-2.