If the length unit is 1nm = 10 (- 9th power), then 25100nm = --- m?

If the length unit is 1nm = 10 (- 9th power), then 25100nm = --- m?

1nm=10＾（－9）m
25100nm=25100*10＾（－9）m=2.51*10^(－5）m
If the length unit is 1nm = 10 (- 9th power), then 25100nm = 2.51 * 10 ^ (- 5) M
The range of function y = x + 8 / X (x is not equal to 0) is
The function y = x + 8 / x x x ≠ 0, x ^ 2-yx + 8 = 0 is obtained by multiplying both sides of the function by X, because the equation has a solution
If △ = y ^ 2-32 ≥ 0, y ≥ 4 √ 2 or Y ≤ - 4 √ 2,
Then the range of the function is (- ∞, - 4 √ 2] ∪ [4 √ 2, + ∞)
In the same base power: A ^ n * a ^ m = a ^ n + m power: (a ^ n) ^ m = a ^ nm product multiplication
In the same base power: A ^ n * a ^ m = a ^ n + m power: (a ^ n) ^ m = a ^ nm product power: (AB) ^ n = a ^ B ^ n, why should n and m be positive integers, not zero and negative numbers? P.S. (* is the multiplication sign)
This is the question of which grade, this should not be zero, but can be negative. You can try with any value
Let X be greater than or equal to 0, y be greater than or equal to 0, and X + 2Y = 1 / 2, find the maximum value of the function p = 8xy + 4Y ^ 2 + 1
P = 4Y (y + 2x) + 1, consider the mean inequality, that is, 4 / 3 * 3Y (y + 2x) is less than or equal to 4 / 3 * (4Y + 2x) ^ 2 / 4 = 1 / 3, so the maximum value is 4 / 3
The eighth power of (- 2A), divided by the second power of [- (2a)]
The eighth power of (- 2A), divided by [- (2a) to the second power]
=-The sixth power of (2a)
=-The sixth power of 64a
-2 times a to the sixth power
The range of the function y equal to the square of x minus two x plus five out of two
y=5/(x²-2x+2)
Let t = x & # 178; - 2x + 2 = (x-1) &# 178; + 1 > = 1
So 0
Fourth power of (P-Q) divided by (Q-P) third power * (P-Q) second power
Fourth power of (P-Q) divided by (Q-P) third power * (P-Q) second power
=-Fourth power of (P-Q) divided by (P-Q) third power * (P-Q) second power
=-4 + 3 + 2 power of (P-Q)
=-The 9th power of (P-Q)
The range of function y = 2 + log2x (x ≥ 1) is______ .
When x ≥ 1, log2x ≥ 0, so y = 2 + log2x ≥ 2, so the range of function y = 2 + log2x (x ≥ 1) is [2, + ∞), so the answer is [2, + ∞)
The third power of x plus the second power of x minus 12 factorization factor
x³+x²-12
=(x³-2x²)+(3x²-6x)+(6x-12)
=x²(x-2)+3x(x-2)+6(x-2)
=(x-2)(x²+3x+6)
The range of the function y = 1 / LNX (x ≥ E) is
（0,1）