A 16 cm long and 8 cm wide rectangular sheet iron, can you cut it into five pieces and weld it into a rectangular container with a square bottom? (no waste) draw a sketch and figure out the volume of the container?

The drawing is as follows: the volume of the container is 8 × 8 × 2 = 128 (cubic centimeter). A: the volume of the container is 128 cubic centimeter

A 16 cm long and 8 cm wide rectangular sheet iron, can you cut it into five pieces and weld it into a rectangular container with a square bottom? (no waste) draw a sketch and figure out the volume of the container?

The drawing is as follows: the volume of the container is 8 × 8 × 2 = 128 (cubic centimeter). A: the volume of the container is 128 cubic centimeter

Below is a rectangular sheet of iron 8 cm long and 4 cm wide. Can you cut it into five pieces and weld it into a rectangular container with a square bottom

Can - be set as square ABCD, midpoint e of AB, midpoint BC, midpoint g of CD, midpoint ad
There are four cutting tools connecting AF, BG, CH and De to form the midpoint of a square in the middle br > each remaining right angled trapezoid and triangle forms a square
5 congruent square

Given the vertex P of parabola y = x & sup2; + 4x + C on the straight line y = 3x + 5, we can find the coordinates of vertex P

Y=X²+4X+C=(x+2)^2-4+C
The vertex coordinates (- 2, C-4) are substituted into the linear equation
C-4=-6+5
C=3
The vertex coordinates are (- 2, - 1)

Why is this proposition wrong: there are two congruent triangles with the opposite angles of two sides and one side (this angle is obtuse angle) corresponding to the same

SSA can't prove congruence
Only ASA AAS SAS SSS HL can prove congruence

Given that the function f (x) = x ^ 3 + ax ^ 2-x + 2 a belongs to real number, if the monotone interval of F (x) is (- 1 / 3,1), find the function y = f (x) image passing through point (1,1)
Given that the function f (x) = x ^ 3 + ax ^ 2-x + 2 a belongs to real number, if the monotone interval of F (x) is (- 1 / 3,1), find the function
Y = f (x) the area of the figure enclosed by the tangent line of the image passing through the point (1,1) and two labels

Y'=3x^2+2ax-1
The extremum points of the function are - 1 / 3 and 1 respectively
So we substitute 1 into the above formula to get a = - 1
So y = x ^ 3-x ^ 2-x + 2
Because the point (1,1) is not on the function
So let the tangent coordinates be (x, y)
Then (Y-1) / (x-1) = 3x ^ 2-2x-1 slope
Take y into the above equation
（x^3-x^2-x+2-1）/(x-1)=3x^2-2x-1
We get x = 0 or 1
When x = 1, there is only one intersection point between the line and the two coordinate axes, so the line is rounded off
So x = 0, so y = 2, and the slope is - 1
So the line is Y-2 = - 1 X
The line intersects two coordinate axes at (0,2) and (2,0)
So s = 0.5 * 2 * 2 = 2 OK

(x+y)^2-9*y^2
Factorization, there is a process on the line

Original formula = (x + y) ^ 2 - (3Y) ^ 2
=(x+y+3y)(x+y-3y)
=(x+4y)(x-2y)

If the value range of X in quadratic function y = - 3x + 1 is 2 ≤ x ≤ 5, the maximum value of this function can be obtained
The function is y = - 3x & # 178; + 1

Because y = - 3x ^ 2 + 1, the axis of symmetry is x = 0, and the opening is downward
When x 0 is a monotone decreasing function
So when 2 ≤ x ≤ 5, y decreases with the increase of X. when x is the minimum, y is the maximum
When x = 2, y = - 3 * 2 * 2 + 1 = - 11 is the maximum

How to do 2 / 3x-1 / 5x = 7 and 2 / 7?
80-9.8) * 0.6-2.1 how to simplify the operation? If not, give the process to do
7 / 8 * [6 / 7 - (1 / 21 + 3 / 7)] this seems to be easy to calculate
2525-12.512.5*3.14/2

(1)2/3X-1/5X=7+2/7
7/15X=51/7
X=765/49
(2)(80-9.8)×0.6-2.1
=70.2×0.6-2.1
=(70+0.2)×0.6－2.1
=42+0.12-2.1
=40.02
(3)7/8×[6/7-(1/21+3/7)]
=7/8×[(6/7－3/7)-1/21]
=7/8×(3/7-1/21)
=7/8×8/21
=1/3
(4)25²-12.5²×3.14/2
=25²－(25/2)²×3.14/2
=25²×(1-3.14/8)
=625×243/400
=6075/16
This is my conclusion after meditation,
If not, please ask, I will try my best to help you solve it~
If you are not satisfied, please understand~

Given the parabola y = x2 and y = (M2-1) x + m2, when m is a real number, there are two intersections between the parabola and the straight line?

If there are two intersections between the parabola y = x & sup2; and the straight line y = (M & sup2; - 1) x + M & sup2;, then the equation
X & sup2; = (M & sup2; - 1) x + M & sup2; has two unequal real roots
When m = 0, the parabola y = x & sup2 and the straight line y = - 1 x have two unequal real roots 0 and - 1
When m ≠ 0, the two equations have two unequal real roots, that is, X & sup2; = (M & sup2; - 1) x + M & sup2; and △ 0
△= b²-4ac=(m²-1)²+4m²＞0
The solution of the inequality is m ∈ R
So when m is any real number, the parabola and the straight line have two different intersections
(I suggest you draw more pictures when you come across such a problem. It's obvious from the pictures.)

A 16 cm long and 8 cm wide rectangular sheet iron, can you cut it into five pieces and weld it into a rectangular container with a square bottom? (no waste) draw a sketch and figure out the volume of the container?