The oil yield of rapeseed is 42%. How many kilograms of oil can be extracted from 2100Kg rapeseed

The oil yield of rapeseed is 42%. How many kilograms of oil can be extracted from 2100Kg rapeseed

2100x42% = 882kg
A: 882kg of oil

The line passing through point P (8,1) intersects hyperbola x2-4y2 = 4 at two points a and B, and P is the midpoint of line ab. the equation of line AB is obtained

Let a (x1, Y1), B (X2, Y2), then X1 + x2 = 16, Y1 + y2 = 2, ∵ x12-4y12 = 4, x22-4y22 = 4, ∵ 16 (x1-x2) - 8 (y1-y2) = 0, ∵ KAB = 2, ∵ the equation of straight line is Y-1 = 2 (X-8), that is, 2x-y-15 = 0

Solution equation: 3x-0.8 = 6 / 5

3x-0.8 = 6 / 5
15x-4=6
15x=10
X = 10 / 15
X = 2 / 3

Let a = {y | y = - x2-2x + 1, X ∈ r}, B = {y | y ≤ t}, and a be truly contained in B, then the value range of real number T is______

A
y=-(x+1)^2+2≤2
A is really contained in B, so t is greater than 2

How to calculate 13.9 * 9.9 + 1.39 simply

13.9*9.9+1.39
=13.9*9.9+13.9*0.1
=13.9*(9.9+0.1)
=13.9*10
=139

Find the linear equation of X + 2y-3 = 0 with respect to the point (- 1, - 3) symmetry

Easy to get (1,1), (5, - 1) on a straight line
The symmetric points of (- 1, - 3) are (- 3, - 7), (- 7, - 5)
Then the line is x + 2Y + 17 = 0

English words 1-31
For example: 1 first 2 second

The 1st First 1st 2nd 2nd 3rd 4th 5th 5th 6th sixth 7th 7th 7th 8th 8th 8th 9th 9th 10th 10th 11th 11th 11th 11th 11th 12th 12th 13th third

Expand the following functions into power series at the specified point, and determine their convergence range (1 + x) in (1 + x), x0 = 0; (x-1) / (x + 1), x0 = 1
(x-1) / (x + 1), and one more question: X / (2-x-x ^ 2), x0 = 0, help me to work it out,

1.f(x) = (1+x) ln(1+x),f '(x) = 1 + ln(1+x),
F '' (x) = 1 / (1 + x) = ∑ n: 0 - > ∞ (- 1) ^ n x ^ n, convergence domain (- 1,1)
Integral: F '(x) = ∑ n: 0 - > ∞ (- 1) ^ n x ^ (n + 1) / (n + 1) = ∑ n: 1 - > ∞ (- 1) ^ (n-1) x ^ n / n
Re integration: F (x) = ∑ n: 1 - > ∞ (- 1) ^ (n-1) x ^ (n + 1) / [n (n + 1)] convergence domain [- 1,1]
2.(x-1)/(x+1) = (x-1) * [1/(2 + x﹣1)],
And 1 / (2 + U) = (1 / 2) / [1 + (U / 2)] = (1 / 2) ∑ n: 0 - > ∞ (- 1 / 2) ^ n u ^ n convergence domain (- 2,2)
The convergence region (- 1,3) is obtained by substituting u = X-1
3.x/(2-x-x^2) = (1/3) x * [ 1/(1-x) + 1/(2+x) ]
Using 1 / (1-x) = ∑ n: 0 - > ∞ x ^ n, 1 / (1 + x) = ∑ n: 0 - > ∞ (- 1) ^ n x ^ n
Do it by substitution or derivation

1 () times () equals () 52

1026×2=2052

Is the negative quadratic power of function y = x an increasing function on (0, + infinity)?

Let's make it clear that it's a minus function
Because the denominator of this formula is the square of X, and y = the square of X, you should know that it increases monotonically on (0, + infinity)
When the denominator increases, the value of the whole function decreases
Composite function