F (x) = x ^ 3 + BX ^ 2 + CX + D (x belongs to R). It is known that f (x) = f (x) - f '(x) is an odd function and f (1) = t (T)

F (x) = x ^ 3 + BX ^ 2 + CX + D (x belongs to R). It is known that f (x) = f (x) - f '(x) is an odd function and f (1) = t (T)

f'(x)=3x²+2bx+c
F (x) = x & sup3; + (B-3) x & sup2; + (c-2b) x + D is an odd function
F(-x)=-F(x)
So B-3 = 0, d = 0
b=3
F(x)=x³+(c-6)x
F(1)=1+c-6=c-5=t
c=t+5
c-6=t-1
F'(x)=3x²+t-1=0
x²=1-t
x=±√[(1-t)/3]
X √ [(1-T) / 3], f '(x) > 0, f (x) increases
√[(1-t)/3]

It is known that the function f (x) = X3 + 3bx2 + CX + D is an increasing function on (- ∞, 0) and a decreasing function on (0, 2), and one root of F (x) = 0 is - B (I) to find the value of C; (II) to prove that f (x) = 0 has real roots X1 and X2 different from - B, and x1, - B and X2 are arithmetic sequences; (III) to find the value range of F (1) if the maximum value of F (x) is less than 16

Fill in the magic square and fill in the nine lattices of the magic square with - 8, - 6, - 4, - 2,0,2,4,6,8, so that all three numbers of the horizontal, vertical, oblique and diagonal are added
Fill - 8, - 6, - 4, - 2,0,2,4,6,8 in the nine lattices of magic square, so that all three numbers of horizontal, vertical, oblique and diagonal are added to 0. If there is a formula to solve the problem, it is better

-6 8 -2
4 0 -4
2 -8 6

To solve the linear equation of three variables (1) 4x + 3x + 2Z = 7 (2) 6x-4y-z = 6 (3) 2x-y + Z = 1

①4x+3y+2z=7
②6x-4y-z=6
③2x-y+z=1
②+③,8x-5y=7----（4）
①-③×2,5y=5,y=1
Substituting y = 1 into (4), 8x-5 = 7, 8x = 12, x = 3 / 2
Substituting y = 1, x = 3 / 2 into 3, 3-1 + Z = 1, z = - 1
x=3/2,y=1,z=-1.

How to do the second question on page 44 of supplementary exercise mathematics (Volume 1, grade 5)

I don't have any supplementary exercises now

If a and B are opposite to each other, C and D are reciprocal to each other, and the square of X is 4, find the value of the algebraic formula a + B-Cd + | X-1 |

From the meaning of the question: a + B = 0, CD = 1, x = 2 or - 2, then when x = 2, a + B-Cd + | X-1 | = 0-1 + 1 = 0; when x = - 2, a + B-Cd + | X-1 | = 0-1 + 3 = 2

6 out of 18 = 3 divided by () = () divided by 3
Don't sleep until you finish

6 out of 18 = 3 divided by (9) = (1) divided by 3

There is a triangle whose bottom is 8 cm and its height is 6 cm. The other one is 10 cm (right triangle). There is a circle in it to find the area of the circle
The answer is 18.0864. How do you find it

8 divided by 2 = 4 S = R2 = 3.14x4x4 = 3.14x16 = 50.24 the answer is definitely right

Solve the equation. 5.5x + 6.7 = 7.8,3.5x-0.8x = 11.34,6.2x-x = 41.6,4 (10-x) = 32,9 (x + 3) = 135,3 (6-x) = 15, (3x + 2) divide by 8 = 7, (80-4x) X10 = 600,9x-14x5.5 = 22

56566

A semicircular vegetable field, 61.68 meters in circumference, how much is the area?
the sooner the better
Understand / be clear! The final answer is 226.08, you are not wrong, please write every step, OK?

Step by step:
Let the diameter of the semicircle be X
x+(1/2)*3.14x=61.68
x+1.57x=61.68
2.57x=61.68
x=24
Since the diameter is 24 meters, the radius is 24 / 2 = 12 meters;
Then the area of the whole circle is calculated by the method of calculating the area of the circle, and divided by 2:
(12 * 12 * 3.14) / 2 = 226.08 square meters