It is known that the volume of a spherical iron block is measured by a rectangular container with a side length of 8 cm and a height of 16 cm It is known that a rectangular container with a side length of 8 cm on its bottom and a height of 16 cm is used to measure the thickness of the container Measure the volume of a spherical iron block. The water in the container is 2 cm away from the mouth of the cup In the container, some water overflows. When the iron block is taken out, the water surface drops by 5cm Product

It is known that the volume of a spherical iron block is measured by a rectangular container with a side length of 8 cm and a height of 16 cm It is known that a rectangular container with a side length of 8 cm on its bottom and a height of 16 cm is used to measure the thickness of the container Measure the volume of a spherical iron block. The water in the container is 2 cm away from the mouth of the cup In the container, some water overflows. When the iron block is taken out, the water surface drops by 5cm Product

It turns out that who is 2 cm away from the mouth of the cup, and the water surface drops 5 cm after taking out the iron ball. In fact, there should be 2 cm left
Centimeter of water should overflow, just because it is water in the beginning, it is not enough to fill the overflow
The original empty 2 cm place. In short, in fact, the water dropped 7 cm!
8×8×7＝448

The volume of a spherical iron block was measured in a cylindrical container with a bottom diameter of 8 cm and a height of 17 cm
When the iron block is put into the container, some water overflows. When the iron block is taken out, the water surface drops 5 cm

r=8/2=4
Square of 3.14 * 4 = 50.24
50.24*5=251.2

A cuboid shaped glass container, measured from the inside, is 60cm long and 40cm wide. There is 20cm water in the container. Now
After immersing the container with a rectangular iron block 50cm long and 30cm wide, the water surface is 25cm away from the bottom of the container?

Volume of iron block = 60 * 40 * (25-20) = 12000
Weight = 12000 * 7.8 = 93600g

The difference between the binomial coefficient of the fourth term and the coefficient of the fourth term, and how to find the difference?

The fourth coefficient of the binomial of (1-2 / 3) ^ n includes the coefficient C (n, R *) * (- 2 / 3) ^ R, that is, t (R + 1) = C (n, R) (- 2 / 3) ^ R, (r = 3), that is, t (3 + 1) = T4 + C (n, 3) (- 2 / 3) ^ 3 = [(n (N1) (n-2) / 1 * 2 * 3] * (- 8 / 27) = - 4N (N1 -) (n-2) / 81. The second coefficient of the fourth term: t (R + 1) = C (n, 3) = n (n-1) (...)

(+ 1) + (- 2) + (- 3) + (- 4)... + (+ 2005) + (- 2006) + 2007?

It's very simple. You count two numbers together
In addition, you may be wrong in the topic. 3 should be positive. Here, odd numbers should be positive and even numbers negative
=（1-2）+（3-4）+（5-6）+······+（2005-2006）+2007
Each bracket is - 1, in which there are 2006 divided by 2 brackets, that is - 1 times 1003
=（-1）*1003+2007
=1004

In the triangle ABC, the equation (1 + x) Sina + 2xsinb + (1-x) sinc = 0 has two unequal real numbers
Find the value of A

The original formula can be transformed into: (Sina sinc) x & sup2; + 2sinbx + Sina sinc = 0, sina ≠ sinc, 4sin & sup2 is obtained from △ > 0; B-4 (Sina sinc) & sup2; > 0, Sin & sup2 is obtained; b > (Sina sinc) & sup2 is obtained; SINB > Sina sinc is obtained

An approximate solution of the equation x ^ 2-2x-5 = 0 in the interval (3,4) is obtained by dichotomy. An approximate solution of the equation x ^ 2-2x-5 = 0 in the interval (3,4) is obtained by dichotomy
Accurate to 0.1

Let f (x) = x ^ 2-2x-5, then f (3) = - 2, f (4) = 3f (3.5) = 0.25F [(3 + 3.5) / 2] = f (3.25) = - 0.9375f [(3.25 + 3.5) / 2] = f (3.375) = - 0.359375f [(3.375 + 3.5) / 2] = f (3.4375) = - 0.05859375f (3.45) = 0.003085937, so f (3.4375) = 0.4375

How to choose free unknowns such as 12-1420-53-7-3000 in linear algebra

1 2 -1 4 2
0 -5 3 -7 -3
0 0 0 0 0
The column of the first nonzero element in the nonzero row corresponds to the constrained unknown quantity
The problem is x1, x2
The rest are free unknowns
The problem is X3, X4

The case of root of 5x square + 8x-1 / 3 = 0

8 ^ 2-4 * 5 / 3 > 0, all have two unequal real roots

Given a = {x | x ＜ - 1 or X ＞ 2}, B = {x | 4x + a ＜ 0}, when B ⊆ a, the value range of real number a is obtained

∵ a = {x | x ＜ - 1 or X ＞ 2}, B = {x | 4x + a ＜ 0} = {x | x ＜ - A4}, ∵ a ⊇ B, ｜ - A4 ≤ - 1, that is, a ≥ 4, so the value range of a is a ≥ 4