A cube box is 2 decimeters long. How many square decimeters is the surface area of the box? How many square decimeters is the area of the trademark pasted around?

1.2 * 2 * 6 = 24 square decimeters
2.2 * 2 * 4 = 16 square decimeters

A cube shaped biscuit box, the edge length is 2 decimeters, how many square decimeters is the surface area of the biscuit box? How many square decimeters is the area of the trademark pasted around it?

2 * 2 * 6 = 24 square decimeters, the trademark is also 24 square decimeters

As shown in the figure, the parabola y = - 3 / 8x2-3 / 4x + 3 intersects with the x-axis at two points a and B (point a is on the left side of point B), and intersects with the y-axis at point C (1) to calculate the coordinates of points a and B; (2) let d be any point on the symmetry axis of the known parabola, and calculate the coordinates of point d when the area of △ ACD is equal to the area of △ ACB; (3) if the line L passes through point E (4,0), M is the moving point on the line L, When there are only two right triangles with vertices a, B and m, the analytic expression of line L is obtained

(1) Let y be equal to 0 and multiply the cross to get the result
A（-4,0）B（2,0）
(2) (- 1,27 / 4) (- 1, - 9 / 4) the process is a little complicated. I will describe the following ideas:
First, calculate the C point of the y-axis intersection of the function image, and convert the analytic expression of the function into y = a (X-H) &# + K form, and use the matching method
Connect AC and BC, calculate s △ ABC and vertex coordinates (- 1,27 / 8)
The analytic expression of AC is calculated so that x = - 1, y is obtained, marked as point E, let D (- 1, m)
Then start the classification:
① Above point E: s △ ACD = 1 / 2 * (m-ye) * 3 + 1 / 2 * 1 (m-ye), M = 27 / 4
② Below point E: s △ ACD = 1 / 2 * (ye-m) * 3 + 1 / 2 * 1 (ye-m), M = - 9 / 4

As shown in the figure, it is known that the line L: y = √ 3 / 3x, the vertical line passing through point a (0,1) as Y-axis intersects line L at point B, the vertical line passing through point B as line L intersects Y-axis at point A1, and the vertical line passing through point A1 intersects Y-axis at point A1
The vertical line of the axis intersects the line L at point B1, and the vertical line passing through point B1 intersects the line y at point A2. If this method continues, the coordinate of point an is

It is easy to know that the angle between the line L and the x-axis is 30 °, by = ay = 1, BX = ay &; √ (3) = √ (3) & nbsp; a1y = BX &; √ (3) + ay = 3 + 1 = 4, & nbsp; b1x = a1y &; √ (3) = 4 √ (3) & nbsp; a2y = b1x &; √ (3) + a1y = 12 + 4 = 16, & nbsp; b2x = a2y &; √ (3) = 16 √ (3) & nbsp

As shown in the figure, the analytical expression of line L1 is: y = - 3x + 3, and L1 and X axis intersect at point D, line L2 passes through points a and B, and lines L1 and L2 intersect at point C
As shown in the figure, the analytical expression of line L1 is: y = - 3x + 3, and L1 and X axis intersect at point D, line L2 passes through points a and B, and lines L1 and L2 intersect at point C
(1) The analytic expression of line L2 is obtained;
(2) Calculate the area of △ ADC;
(3) There is another point P on line L2 which is different from point C, so that the area of △ ADP and △ ADC is equal, and the coordinates of point P are obtained;
(4) The expression of line L2
emergency

(1) let y = 0, that is, - 3x + 3 = 0, then x = 1, D (1,0); let L2: y = KX + B, through a (4,0) and B (3, - 3 / 2), then the system of equations: 0 = 4K + B, - 3 / 2 = 3K + B, the solution of which is k = 3 / 2, B = - 6, the analytic formula of which is y = 3 / 2x-6; (2) the solution of the system of equations: y = - 3x + 3Y = 3 / 2x-6, then x = 2, y = - 3, - C (2, - 3), ad = 2, - s Δ ADC = 1 / 2 * a

A straight line passing through the point (1, x) intersects with the positive half axis of X axis and Y axis respectively at two points a and B. o is the origin. If the area of triangle is the smallest, the linear equation is

I don't know. Is it the largest or the smallest area

It is known that the line L and the two coordinate axes intersect at points a and B respectively, and the area of the triangle is 8. The length of AB is equal to twice the distance from the origin o to the line L, then the square of the line L

AB=2d
Then area = 2D * D △ 2 = 8
d=2√2
And the center line on the hypotenuse is half of the hypotenuse, and it's half here
So it's isosceles
So x + y + a = 0 or X-Y + a = 0
d=|0+0+a|/√(1+1)=2√2
a=±4
therefore
x+y+4=0
x+y-4=0
x-y+4=0
x-y-4=0

The line L passing through point P (2,3 / 2) intersects with the positive half axis of X axis and the positive half axis of Y axis at a, B and O respectively, and the area of △ AOB is 6
The equation for finding the line L~
If possible + the way to solve this problem~

3X+4Y=12

When the area s of △ AOB (o is the origin) is the smallest, the line L passing through point P (2,1) intersects the positive half axis of X axis and the positive half axis of Y axis at points a and B respectively,
Make a straight line L through point P (2,1), intersect the positive half axis of X axis and the positive half axis of Y axis at points a and B respectively. When the area s of △ AOB (o is the origin) is the smallest, find the equation of the straight line L and the minimum value of S

Let the complement of the inclination angle be a
Draw a vertical line through (2,1) axis
The longitudinal intercept is 1 + 2tana
The cross section is 2 + 1 / Tana
So:
S=0.5(1+2tana)(2+1/tana)
0 = 2 + 2 = 4 (expansion, using inequality)
If and only if 4tana = 1 / Tana, Tana = 0.5, the minimum value of S is 4, then l: x + 2y-4 = 0

The intersection of the image of the linear function y = - N / (n + 1) * x + 1 / (n + 1) (n is a positive integer) and the x-axis, Y-axis is a, B, O as the origin, let the triangle AOB area be SN
(1) Seek S1
(2) Find S1 + S2 + S3 + S4 + +S2012

The point of intersection with X axis is (1 / N, 0) and Y axis is (0,1 / (n + 1)); Sn = 1 / 2 × 1 / N × 1 / (n + 1) = 1 / 2 × (1 / n-1 / (n + 1)) (1) S1 = 1 / 2 × (1-1 / 2) = 1 / 4 (2) S1 + S2 +... + s2012 = 1 / 2 × (1-1 / 2 + 1 / 2-1 / 3 +... + 1 / 2012-1 / 2013) = 1 / 2 × 2012 / 2013 = 1006 / 2013

A cube box is 2 decimeters long. How many square decimeters is the surface area of the box? How many square decimeters is the area of the trademark pasted around?