The length of a rectangular steel plate is three times of its width. The width is known to be 3 * 10 cm to the third power. Find the area of this rectangular steel plate. (the result is expressed by scientific counting method.)

The length of a rectangular steel plate is three times of its width. The width is known to be 3 * 10 cm to the third power. Find the area of this rectangular steel plate. (the result is expressed by scientific counting method.)

Width: 3 * 10 ^ 3cm = 30m
Length: 3 * 3 * 10 ^ 3 = 90m
Area: 30 * 90m ^ 2 = 2.7 * 10 ^ 3M ^ 2

It is known that a, B and C are three real numbers which are not all zero, then the root of the equation x2 + (a + B + C) x + A2 + B2 + C2 = 0 is ()
A. There are two negative roots B. There are two positive roots C. There are two positive roots and one negative root d. There are no real roots

∵ △ = (a + B + C) 2-4 (A2 + B2 + C2) = - 3a2-3b2-3c2 + 2Ab + 2BC + 2Ac = - (A-C) 2 - (B-C) 2 - (a-b) 2-a2-b2-c2, and a, B and C are three real numbers which are not all zero, ∵ (A-C) 2 - (B-C) 2 - (a-b) 2 - ≤ 0, a2-b2-c2 < 0, ∵ △ < 0, ∵ the original equation has no real root, so D

If the absolute value of M is equal to negative m, then M is a number. If M is equal to negative m, then M is a number. If M is equal to negative m, then M is a number
Please answer, the teacher hand copied the paper, is it wrong

、、、
The absolute value of M is negative m, because the absolute value can only be positive and 0, so m is 0 or negative
If the absolute value of M is m, then M is 0 or positive
If M is equal to negative m, it's very simple. M can only be zero
If M is greater than negative, M is positive
If M is less than negative, M is negative
No, I can't. OK, give me a compliment

The function f (x) = XX (ax-3) defined on R is known, where a is a constant. If x = 1 is an extreme point of the function, find the value of A

f(x)=xx(ax-3)
f'(x)=2x(ax-3)+ax^2
=3ax^2-6x
X = 1 is an extreme point of the function = = > F '(1) = 0
3a-6=0
a=2

The minimum value of | ab | is______ ．

∵ x2 + y2 = 4's center O (0, 0), radius r = 2, the distance d from the point (0, 1) to the center O (0, 0) is 1, the point (0, 1) is inside the circle x2 + y2 = 4, as shown in the figure, | ab | is the smallest, the chord center distance is the largest 1, | ab | min = 222 − 1 = 23

Univariate quadratic function y = x ^ 2-mx + m-2, there are two intersections (x1,0), (x2,0) between image and x-axis, try to use m to represent X1 ^ 2 + x2 ^ 2 and find its minimum value
Specific process, thank you!

Let y = 0x ^ 2-mx + m-2 = 0. According to WIDA's theorem, X1 + x2 = mx1x2 = m-2x1 ^ 2 + x2 ^ 2 = X1 ^ 2 + x2 ^ 2 + 2x1x2-2x1x2 = (x1 + x2) ^ 2-2x1x2 = m ^ 2-2 (m-2) = m ^ 2-2m + 4 = m ^ 2-2m + 1 + 3 = (m-1) ^ 2 + 3. Because delta = m ^ 2-4 (m-2) > 0, the range of m ^ 2-4m + 8 > 0 (m-2) ^ 2 + 4 > 0m is r, so for

Monotone decreasing interval of absolute value of function y = log2 X-1

1) If your topic is:
y=log2(|x-1|)
The original function can be divided into:
y=log2(t) (t>0)
t=|x-1|
Y = log2 (T) increases monotonically when t > 0, from t > 0 = = > | X-1 | 0, that is, X ≠ 1
When x

Let f (x) defined on R satisfy f (x) times f (x + 2) = 13, if f (1) = 2, find f (99)
I know it's a periodic function, but I'll substitute x = 1 357 and find out how it's a function with a period of 2. In that case, the answer is not 2?
F1 = 2 F3 = 13 / 2 F5 = 2 F7 = 13 / 2. What's wrong?

Because f (x) f (x + 2) = f (x + 2) f (x + 4) = 13
So f (x) = f (x + 4)
So the period of F (x) is 4
F (3) = 13 / 2
So f (99) = f (95) =. = f (3) = 13 / 2

It is known that the absolute value of X is less than or equal to 1, and the absolute value of Y is less than or equal to 1. The analysis method proves that the absolute value of x plus y is less than or equal to 1 plus XY!

|X + y | & sup2; ≤ | 1 + XY | & sup2;, expand, namely prove: X & sup2; + Y & sup2; ≤ 1 + X & sup2; Y & sup2;, namely prove: X & sup2; (1-y & sup2;) - (1-y & sup2;) ≤ 0, namely: (1-y & sup2;) (1-x & sup2;) ≥ 0

When and where to do this simple sentence as the subject, predicate with singular or plural

Where to go and what to do have not been decided