Ninth grade math problems

Ninth grade math problems, detailed process, urgency, online solution, etc. ;

If o is OC⊥AB, then:

AC=CB=AB/2=(√3)/2

So: OC= 1/2 is obtained by Pythagorean theorem.

So: ∠ AOC = 60.

So: ∠ AOB = 120.

So: the area of the sector is (π/360) *120 = π/3-the answer to the first question.

Therefore, the length of the lower arc AB is (2π/360)* 120=2π/3.

So: the circumference of the sector is 2+(2 π/3)-the answer to the second question.

Ninth grade math problem, detailed process, (1) The profit per loaf is (x-0.5) yuan,

The quantity of bread sold is160-20 (x-0.7)/0.1= 300-200x.

(2)y=(x-0.5)(300-200x)。

(3)y=-200x^2+400x- 150

=-200(x- 1)^2+50，

When the price of bread is 1 yuan, the daily sales profit of this store is the largest, and it is 50 yuan.

( 1)f(x)=-x^2+2(m- 1)x+m+2

According to known f (0) >: 0, m+2 >: 0 get m & gt-2.

Let a be (a, 0) a.

Vieta Theorem 2 (m- 1) =-4a. -(m+2) =-5a 2。 A =- 1。 (A = 3/5 >； 0 give up)

So there is m=3.

So the parabolic equation is y =-x 2+4x+5.

Ninth grade math problems, online, etc. , there should be a detailed process. Don't add conditions upstairs without authorization.

1. Connect OF, OD Because CD is the chord of the circle O perpendicular to OB, DM=MC is obtained according to the symmetry of the circle.

So the center of the circle with a diameter of CD is m, so EM=MC.

Because the angle AMC is 90 degrees, the angle FCD=45 degrees.

So the central angle FOD=90 degrees, so FD square =OF square +OD square = 1+ 1=2.

So FD= root number 2

2. What we require is the area of quadrilateral AEFD. The known areas are △BEF, △CDF and △BCF, which can be divided and summed. Because △ABF and △ADF are the same height, △AEF and △ACF are the same height, it is easy to get the equation.

(5+x)/y= 10/8

x/(y+8)=5/ 10

Solving the equation gives x= 10 and y= 12, so the area of the quadrilateral AEFD is 22.

The ninth grade math problem needs a detailed process. The two images intersect at point P, and the function values are the same, so 2X-6=-X+3, and the solution is X=3. If X=3 is substituted into Y=-X+3 and Y=0, then the coordinate of point P is (3,0).

The detailed process of solving math problems in the ninth grade is 1, and y=(x- 1)? -4

A(- 1，0)、B(3，0)、D(0，-4)

AB=4

( 1)sδABD = AB * OD/2 = 8

(2)sδABP = AB * yp/2 = 4

(i)yp=2

2=(xp- 1)？ -4

xp= 1 √6

p( 1-√6，2)、p( 1+√6，2)

(2) yp=-2

-2=(xp- 1)？ -4

xp= 1 √2

p( 1-√2，-2)、p( 1+√2，-2)

2、a？ -3a+2≥0

b+ 1≥0

c+3≥0

Then a=2 or a= 1 b=- 1 c=-3.

( 1) 2x？ -x-3=0

(2x-3)(x+ 1)=0 x=3/2 or x=- 1.

(2) x？ -x-3=0 x = ( 1 √ 13)/2。

Ninth grade mathematics, seeking the detailed process, wind, stars, stars, legs, stars, weeks, electricity, this is also Chen Xing's star account.

According to Pythagorean theorem, diagonal AC = BD = √ (12+12) = √ 2.

So in the second question, we can know that BO=CO=AO=DO, and then we can get Bo 2+Ao 2 =12 from Pythagorean theorem, so we can get AO=BO=CO=DO=√2/2, so we can know the coordinates.

The detailed process of solving the ninth grade mathematical solution;

(1) Let AM⊥BC be at point M and intersect DG at point N.

∫S△ABC = 30，BC= 10

∴AM=6

∫DG‖BC

∴△ADG∽△ABC

∴DG/BC=AN/AM

∴DG/ 10=(6-x)/6

∴GD=5/3(6-x)

∴y=xDG=-(5/3)x^2+ 10x

According to the domain formula of y = ax 2+bx+C, y ≥-(b 2+4ac)/4a.

The domain is y ≥-(102+4 * (5/3 * 0))/4 (-5/3) =15.