How many square meters is 113000 mu

1 mu = 10000 / 15 = 2000 / 3 = 666.6667 square meters, if you use a calculator to calculate, I will not

How many square meters is an acre

1 ha = 10000 M2 = 100 mu = 15 mu, 1 mu = 100 m2, 1 m2 = 0.0015 mu, 1 mu = 666.67 m2, replace M2 with Mu --- the formula is "add half and move left three". 1 m2 = 0.0015 mu, if 128 M2 is equal to how many mu? The calculation method is to first use 128 plus half of 128: 128 + 64 = 192, and then move the decimal point left three places, If you want to calculate how many square meters 24.6 Mu is equal to, 24.6 △ 3 = 8.2, after 8.2 is doubled, it is 16.4, and then move the decimal point to the right 3 places, that is, the square meter is 16400

We know that the square of x plus the square of Y minus 4x plus y plus 4 plus 1 / 4 is equal to 0, and we can find the value of - x plus 3xy of Y. The Gods help us

∵ x + y-4x + y + 4 + 1 / 4 = (x-4x + 4) + (y + y + 1 / 4) = (X-2) + (y + 1 / 2) = 0, if the sum of two non negative numbers is 0, then both numbers are 0 ∵ X-2 = 0, y + 1 / 2 = 0 ∵ x = 2, y = - 1 / 2 ∵ original formula = (- 1 / 2) ^ (- 2) + 3 × 2 × (- 1 / 2) = 4-3 = 1

1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 +. Express the operation rule with the algebraic expression of n

It seems that this question just let you express the law of operation, so the above answers are not right!
Maybe: 1 / 1 of 2 + 1 / 2 of 2 + 1 / 2 of 2 + 1 / 3 of 2 + 1 / 4 of 2

1,3 / 4,5 / 9,7 / 16
How much is the area after folding the square paper with side length 1 N times

1,3 / 4,5 / 9,7 / 16
How much is the area after folding the square paper with side length 1 N times
The area is halved, so it is the nth power of (1 / 2) ^ n half

Express 999.. 9 (n 9S) with algebraic expression containing n

10^n-1

Expressed by an algebraic expression containing N, the first number of the nth row is (), and the last number is (). There are () numbers in the nth row
I am going to have a class
First line 1
Second line 234
Third line 56789
The fourth line 10, 11, 12, 13, 14, 15, 16,
The fifth line 17.18.19.20.24.22.23.24.25
The sixth line 26.27.28.29.30.31.32.33.34.35.36
The seventh line. 37.38.39.40.41.42.43.44.45.46.47.48.49
The eighth line. 50.51.52.53.54.55.56.57.58.59.60.61.62.63.64

1、(n-1)^2+1
2、n^2
3、2n-1

On the equation K square x square + (2k-1) x + 1 = 0 of X has real roots, find the range of K

K²X²+（2K-1）X+1＝0
△＝（2K-1）²-4K²＝4K²-4K+1-4K²＝-4K+1
When △≥ 0, the equation has real roots
Then - 4K + 1 ≥ 0
K≤1/4
When k = 0, the equation is one variable linear equation
That is, when k ≤ 1 / 4 and K ≠ 0, the quadratic equation with one variable has real roots

If K is a real number, then the root of the equation x2 + (2k + 1) x + k-1 = 0 is ()
A. There are two unequal real roots B. There are two equal real roots C. There are no real roots D. It is impossible to judge the root

∵ a = 1, B = 2K + 1, C = k-1, ∵ △ = b2-4ac = (2k + 1) 2-4 × 1 × (k-1) = 4k2 + 4K + 1-4k + 4 = 4k2 + 5 ＞ 0, the equation has two unequal real roots

Please explain that no matter what the value of K is, the square of the equation K - (2k + 1) x + 4 (k-1 / 2) = 0 always has a real root
In the following bracket is k minus one half

If the equation has a real root, b-square-4ac is greater than or equal to 0. However, it seems that it can not be solved. Is there a mistake

How many square meters is 113000 mu