How to calculate the number after the square,

How to calculate the number after the square,

The number before 16900 square is: 130, that is: the square of 130 is equal to 16900, the number before 136 square is: 2 times the root sign 34

The edge length of a cube is 6 decimeters, its volume is () cubic decimeters, and its surface area is () square centimeters
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216 216 both

If there is no solution to the system of equations ax + 3Y = 9 about X, y, the value of real number a 2x-y = 1 is obtained
If the system of equations ax + 3Y = 9, 2x-y = 1 about X, y has no solution, the value of real number a 2x-y = 1 is obtained

Two equations 2 times 3, get ax + 3Y = 9, 6x-3y = 3; add the two to get (a + 6) x = 12, then x = 12 / (a + 6), no solution, that means the denominator is 0, meaningless. So a + 6 = 0, get a = - 6

Is the absolute value of the sum of two positive numbers equal to the sum of the absolute value of the addend?

The absolute value of the sum of two positive numbers is equal to the absolute value of the sum of addends
It is correct that the absolute value of the sum of two negative numbers equals the sum of their absolute values, because the sum of two negative numbers will become the sum of two positive numbers after extracting a negative sign. Then, the absolute value is similar to the conclusion of positive numbers above

If the derivative of the function f (x) = x + BX & nbsp; & nbsp; (B ∈ R) has zeros in the interval (1,2), then f (x) monotonically increases ()
A. （-2，0）B. （0，1）C. （1，+∞）D. （-∞，-2）

∵ function f (x) = x + BX & nbsp; & nbsp; (B ∈ R) ∵ F & nbsp; '(x) = 1 − bx2 ∵ function f (x) = x + BX & nbsp; & nbsp; (B ∈ R) has zero point in the interval (1,2) ∵ when 1 − bx2 = 0, B = X2, X ∈ (1,2) ∵ B ∈ (1,4) Let f' (x) > 0 & nbsp; get x ＜− B or X ＞

A is less than 0, B is less than 0, then a + B () 0, find the absolute value of a + B is equal to

a+b＜0
Absolute value of a + B = - (a + b) = - a-b
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It is known that y = f (x) is an odd function defined on R with period 4, and when 0 ≤ x ≤ 2, f (x) = x2-2x, then when 10 ≤ x ≤ 12, f (x)=______ ．

∵ y = f (x) is a periodic function with period 4 defined on R, f (X-12) = f (x) from 10 ≤ x ≤ 12, - 2 ≤ X-12 ≤ 0, 0 ≤ 12-x ≤ 2 ∵ 0 ≤ x ≤ 2, f (x) = x2-2x ∵ f (12-x) = (12-x) 2-2 (12-x) = x2-22x + 120, and ∵ y = f (x) is an odd function defined on R, ∵ f (X-12) = - f (12-x) = - x2 + 22x-120, so the answer is: - x2 + 22x-120

The derivative function f (x) = 1 / 2x ^ 2-alnx
To prove that if x > 1, f (x) - 2 / 3x ^ 3 + (a + 1) LNX

Let g (x) = f (x) - (2 / 3) x & # 179; + (a + 1) LNX
Namely:
g(x)=－(2/3)x³＋(1/2)x²＋lnx
The results are as follows
g'(x)=－2x²＋x＋(1/x)=[－2x³＋x²＋1]/(x)=[－(x－1)(2x²＋x＋1)]/(x)
When x > 1, when there is always: G '(x) 1, the maximum value is: G (1) 1, there is: G (x)

Given the function f (x-1) = 2x + 5, then f (x ^ 2) =?
f(x-1)=2x+5=2(x-1)+7
f(x)=2x+7
f(x^2)=2x^2+7
How did that 5 become a 7
But even if we calculate 2x + 5 as 2 (x-1) + 5, it is only 2x + 3. How can it be + 7

Let me tell you this: 2x + 5 = 2x-2 + 7, right? So it's equal to 2 (x-1) + 7

The value of M, n can be obtained by dividing the negative quadratic power of (n + 1 power of a, m power of B) by the n power of a, and B = the negative 5th power of a, and the negative 3rd power of B

So - 2 (n + 1) - n = - 5;
-2n-2-n=5;
3n=-7;
n=-7/3;
-2m-1=-3;
2m=2;
m=1;
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