A. The bottom area ratio of two cylindrical containers B is 4:3. The water depth of container a is 20 cm, and that of container B is 10 cm. Now, the same amount of water is injected into the two containers to make the water depth of the containers equal. At this time, the water depth of container a is 20 cm, and that of container B is 10 cm______ Cm

A. The bottom area ratio of two cylindrical containers B is 4:3. The water depth of container a is 20 cm, and that of container B is 10 cm. Now, the same amount of water is injected into the two containers to make the water depth of the containers equal. At this time, the water depth of container a is 20 cm, and that of container B is 10 cm______ Cm

Let the water depth be xcm, then the bottom area of (x-20) × a = (X-10) × B, that is, the area of X − 20x − 10 = & nbsp; B, the area of a = 34,4 × (x-20) = 3 × (X-10), & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 4x-80 = 3x-30, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 50; a: the water depth is 50cm

There are two cylindrical containers a and B, the bottom area is 5:3, the water depth of container a is 20 cm, and that of container B is 10 cm
The bottom area of the two cylindrical containers is 5:3. The water depth of container a is 20 cm, and that of container B is 10 cm. Add the same amount of water to the two containers. The water depth of the two containers is equal. How many cm is the water depth at this time?

The water depth after adding water is set as HCM
(h-20)*5=(h-10)*3
h=35cm

Decomposition factor: (1) 75X ^ 2-48y ^ 2 (2) (a-b) (X-Y) - (B-A) (x + y) (3) a ^ 2 + 4AB + 4B ^ 2 (4) 3ax ^ 3-108ax (5) 81a ^ 4-1

Decomposition factor: (1) 75X ^ 2-48y ^ 2 = 3 (25X ^ 2-16y ^ 2) = 3 (5x + 4Y) (5x-4y) (2) (a-b) (X-Y) - (B-A) (x + y) = (a-b) (X-Y) + (a-b) (x + y) = (a-b) (X-Y + X + y) = 2x (a-b) (3) a ^ 2 + 4AB + 4B ^ 2 = (a + 2b) ^ 2 (4) 3ax ^ 3-108ax = 3ax (x ^ 2-36) = 3ax (x + 6) (X-6) (5) 81a ^ 4-1 = (9

It is known that the module of vector a = the module of vector b = 2, and the angle between vector a and vector B is 60
Find the angle between vector a + vector B and vector a, and the angle between vector a - vector B and vector a

In this paper, A-B |a-b |a-b |a-b | A-B | A-B |a-b |a-b |a-b | A-B |b ||a |a-b | B | A-B | B | B | B | B | B | a + B | a + B | B 124\\\| a + B | a | a + B | a + B | a + B | a + B | a + B \- B | x | a | = (4-2) / 2 × 2 = 1 / 2; = 60 °; if this

As shown in the figure, in △ ABC, be and CD are bisectors of ∠ ABC and ∠ ACB respectively. An ⊥ be is in N, am ⊥ CD is in M. verify Mn ∥ BC

Extend an and am to BC and PQ respectively
An ⊥ be and be is bisector of ∠ ABC = > BAP is isosceles triangle = > n is midpoint of AP
Similarly, M is the midpoint of AQ, according to the median theorem Mn / / BC

Factorization of 2x to the second power + 4xy-6ax + 3a-x-2y

2X + 4xy-6ax + 3a-x-2y
＝（2x^2-x）+(4xy-2y)-(6ax-3a)
=2x(2x-1)+2y(2x-1)-6a(2x-1)
=(2x+2y-6a)(2x-1)
=2(x+y-3a)(2x-1)

ρ=xcosθ+ysinθ
What is the meaning of ρ in the linear polar coordinate equation and how does the formula come from?

The polar axis is the distance between the points and poles

As shown in the figure, ab = AC, ad ⊥ BC at point D, ad = AE, AB bisects ∠ DAE and intersects de at point F. please write out three pairs of congruent triangles in the figure and select one pair to prove it

(1) (2) taking △ ADB ≌ ADC as an example, it is proved that ≌ ad ⊥ BC, ≌ △ AFE, ≌ BFD ≌ △ BFE, ≌ Abe ≌ △ ACD = 90 °. In RT △ ADB and RT △ ADC, ab = Acad = ad ≌

Grade 8 Volume 1 conversion of physical units
12.5mm=1.25*?km
10nm=1.0*?cm=1.0*?m
I admit that I am very happy
But really dizzy, said to go in, but, good dizzy

First, find out the - 6 km of 1 mm = 10
12.5mm=12.5*1mm
=12.5 * 10 - 6 km
=1.25 * 10 - 5 km
According to this method step by step, this is actually the method of physical unit conversion
You can practice it

Given the function f (x) = x + 2, X is less than or equal to - 1, the square of X, - 1 is less than x, less than 2, 2x, X is greater than or equal to 2, find f (- 4), f (- 3)
The value of F [f (- 2)] (2) f (a) = 10, find the value of A

1, - 4, - 3 and - 2 are all less than - 1, so they all need to be brought into f (x) = x + 2 for calculation, so f (- 4) = - 4 + 2 = - 2, f (- 3) = - 3 + 2 = - 1F (- 2) = - 2 + 2 = 0, so f [f (- 2)] = f (0) is between - 1 and 2, so f [f (- 2)] = f (0) = x ^ 2 = 02, let x + 2 = 10, (1) x ^ 2 = 10, (...)