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A and B are two cuboid containers. Container a is 40 cm long, 30 cm wide and 24 cm high. Container B is 20 cm long, 20 cm wide and 36 cm deep. Pour the water in container B into container a until the water depth in the two containers is the same. How many mm is the water depth in both containers?
Bottom area of a container: 40 × 30 = 1200 (square centimeter)
Volume of water in container B: 20 × 20 × 36 = 14400 (cm3)
Bottom area of container B: 20 × 20 = 400 (square centimeter)
Because the bottom area of container a is three times that of container B: 1200 △ 400
Therefore, when the water depth in the two containers is the same, the volume of water in container a is three times of that in container B
Volume of remaining water in container B: 14400 ÷ (3 + 1) = 3600 (cm3)
Water depth: 3600 △ 400 = 9 (CM)
Party A's container is 40 cm long and 25 cm wide, and Party B's cuboid container is 30 cm long and 20 cm wide. At this time, the water depth in Party B is 20 cm
If you pour part of B's water to a so that the height of the two containers is the same, how many centimeters is the water depth in B?
Water = 30 × 20 × 20 = 12000 CC
Present depth = 12000 ^ (40 × 25 + 30 × 20) = 7.5cm
The water depth of container a is 5cm long 40 wide 30b the water depth of container B is 23cm long 30 wide 20. Pour part of the water in container B into a to make the water height of two containers equal. How many centimeters is the water depth at this time
40*30*5+30*20*23=(40*30+30*20)*h
h=11(cm)
What kind of geometry will the shadow turn around the left line? What's the volume of the shadow
It's a right triangle. On a rectangle, the length of the rectangle is 4cm, the width is 3cm, and the height of the triangle is 9cm
Although I don't know what kind of figure you describe, I guess it's a cone with a cylinder under it
Pi = 3.14
Cone H1 = 9-4 = 5cm, bottom radius r = 3cm
Cylinder r = 3cm, h2 = 4cm
Volume = 1 / 3 * pi * 3 * H1 + pi * 3 * 3 * h2 = 15 * PI + 36 * pi = 51 * PI cm3
The square of 2a is 2 times a times a
Wrong
2 times a times a
2×a×a=2a²
The square of 2a is
（2a）²=4a²
2a²≠4a²
So wrong
The known vectors a = (sin C, 1), B = (1, cos C), - π / 2 < C < π / 2
(1) If a ⊥ B, find C (C is the angle name)
(2) Finding the maximum of | a + B |
(1)
If a ⊥ B, A.B = 0, sinc + COSC = 0
=> sin(c+π/4)=0
=>C + π / 4 = k π, (k is an integer)
=>c=kπ-π/4
And - π / 2 < C < π / 2
=>C = - π / 4 (when k is 0)
(2)
a+b=(sinc+1,1+cosc)
|a+b|^2=(sinc+1)^2+(cosc+1)^2=3+2(sinc+cosc)=3+2√2sin(c+π/4)
Looking at sin (c + π / 4), the range of C is - π / 2 < C < π / 2
So the range of C + π / 4 is - π / 4
As shown in the figure, the area of triangle ABC is 16, D is the midpoint of AC, and E is the midpoint of BD. what is the area of quadrilateral cdef?
Because D is the mid point of AC, so s △ abd = s △ BDC = 16 △ 2 = 8, because e is the mid point of BD, so s △ abd = s △ AED = 8 △ 2 = 4, s △ BEC = s △ Dec = 8 △ 2 = 4, because D is the midpoint of AC, so s △ abd = s △ BDC = 16 △ 2 = 8, because e is the mid point of BD, so s △ abd = s △ BDC = 16 △ 2 = 8, because e is the mid point of BD, so s △ Abe = s △ Abe = s △ AED = s △ AED = s △ AED = 8 △ AED: s △ AED = 8 △ AED: s △ bid: s △ CEF: s △ CEF: as (4 + 4 + 4 + 4 + 4 + 4 (4 + 4 + 4 + 4) is (4 + & nbsp & AMP & nbsp; & nbsp; & nbsp nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; &Nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 12x = 32, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & x = 83, so the area of quadrilateral cdef is: 83 + 4 = 203; a: 203
Function image drawing
How to draw an image like y = | 1 + 2x | + | 2-x |
This is a piecewise function
When x ≤ - 1 / 2, y = - (1 + 2x) + (2-x) = - 3x + 1
When - 1 / 2 < x < 2, y = 1 + 2x + 2-x = x + 3
When 2 ≤ x, y = 1 + 2x + X-2 = 3x-1
Then you can draw the image of this region according to the range of X in the coordinate system
Good study
Space vector math problem!
The cube abcd-a'b'c'd ', grows to 3, and points E.E'. G.h.f
(1) Find the distance from point B to EGH
(2) Calculate the distance between EF and CD
(3) Find the distance from AB to plane a 'B' C 'D
(4) Finding: plane EFG and plane e'b'h
Which side is the trisection point of the side where the point E.E '. G.h.f is located
As shown in the figure, in △ ABC, ∠ a = 36 °, ABC = 40 °, be bisection ∠ ABC, ∠ e = 18 °, CE bisection ∠ ACD? Why?
The reasons are as follows: ∵ a = 36 °, ABC = 40 °, ∵ BCA = 104 °, ∵ ACD = 76 °. ∵ be = ABC, ∵ CBE = 20 °, ∵ e = 18 °, ∵ BCE = 142 °, ∵ ECA = 38 °, ∵ ECD = 38 °,
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