1 / 10 times the absolute value of 2x-1 + the square of 2 (y + 3) = 0, find the cube of X + the cube of Y

1 / 10 times the absolute value of 2x-1 + the square of 2 (y + 3) = 0, find the cube of X + the cube of Y

1/10*|2x-1|+2(y+3)²=0
So we can get 2x-1 = 0, y + 3 = 0
We can get x = 1 / 2, y = - 3
Cube of X + cube of y = 1 / 8-27
=-215/8

What is the derivative of one in X

One of X is to the power of X - 1, and its derivative is - 1 * x ^ (- 2)

By simple calculation, 49 × 49-51 + 49 × 51

49×49-51+49×51
=49×49+49×51-51
=49×（49+51）-51
=4900-51
=4849

Reduction ratio, 36:96,5 / 4:7 / 3 0.15:9 3.4:1 / 4 6.5:13 27:54 0.35:7 2 / 3:3 / 4

36:96=3：8
5/4:7/3=5/4*3/7=15：28
0.15:9=3/20*1/9=1：60
3.4:1/4 =17/5*4=68：5
6.5:13=13/2*1/13=1：2
27:54 =1：2
0.35:7 =1/20
2/3:3/4=2/3*4/3=8：9

Given that the function g (x) = - x2-3, f (x) is a quadratic function, when x ∈ [- 1,2], the minimum value of F (x) is 1, and f (x) + G (x) is an odd function, the analytic expression of F (x) is obtained

Let f (x) = AX2 + BX + C (a ≠ 0), then f (x) + G (x) = (A-1) x2 + BX + C-3, ∵ f (x) + G (x) is an odd function, ∵ a = 1, C = 3 ∵ f (x) = x2 + BX + 3, symmetry axis X = - B2, when - B2 > 2, that is, B < - 4, f (x) is a decreasing function on [- 1, 2], and the minimum value of ∵ f (x) is f (2

How to calculate 1500 × 120 △ 25 and 3 / 4 × 101-0.75 (write calculation steps)

1500×120÷25
=1500×120÷100×4
=(1500÷100×4)×120
=60×12
=720
3×101-0.75
=3×(100+1)-0.75
=3×100+3-0.75
=300+2.25
=302.25

Why LIM (n →∞) NX ^ n = 0 (where | x | 1) is the limit of higher mathematics

Given that x = - 1 is a root of the equation 2x2 + ax-a2 = 0 about X, then a=______ The other one is______ ．

According to the meaning: 2-a-a2 = 0, solution: a = - 2 or 1, when a = - 2, then 2x2 + ax-a2 = 0 is: 2x2-2x-4 = 0, sorting out: x2-x-2 = 0, (x + 1) (X-2) = 0, solution: X1 = - 1, X2 = 2, another root is: 2, when a = 1, then 2x2 + ax-a2 = 0 is: 2x2 + X-1 = 0, (x + 1) (X-12) = 0, solution

How to calculate 2-9 / 19-10 / 19 in a simple way?

2 - 9/19 - 10/19 = 2 - (9/19 +10/19) = 2 - 19/19 = 2 - 1 =1

Lim I / N (sin π / N + sin 2 π / N +. + sin n π / N) n tends to be positive infinity

cosπ/n(sinπ/n+sin2π/n+.+sinnπ/n)=1/2(sin0+sin2π/n+sinπ/n+sin3π/n+sin2π/n+sin4π/n+.
+sinπ(n-1)/n+sin(n+1)π/n
=1/2[2*(sinπ/n+sin2π/n+.+sinnπ/n)-sinπ/n+sin0-sinπ*n/n+sin(n+1)π/n]
=(sinπ/n+sin2π/n+.+sinnπ/n)+1/2(-sinπ/n+sin(n+1)π/n)
Let (sin π / N + sin2 π / N +. + sin π / N) = M
M*cosπ/n=M+1/2*(-sinπ/n+sin(n+1)π/n)
M=(sinπ/n-sin(n+1)π/n)/2(1-cosπ/n)=(sinπ/n)/(1-cosπ/n)
=[2*sin(π/2n)*cos(π/2n)]/[2sin^2(π/2n)]=cos(π/2n)/sin(π/2n)
lim 1/n(sinπ/n+sin2π/n+.+sinnπ/n)=limM/n=cos(π/2n)/[n*sin(π/2n)]
Let 1 / N = x and N tend to be positive infinity, then x → 0+
limM/n=lim(x→0+)cos(π*x/2)*x/sin(π*x/2)=lim(x→0+)x/sin(π*x/2)
Because when x → 0 +, sin (π * x / 2) and π * x / 2 are infinitesimal
So the above formula = LIM (x → 0 +) x / (π * x / 2) = 2 / π