I want people to teach the press, the oral problem is the fraction addition and subtraction method

I want people to teach the press, the oral problem is the fraction addition and subtraction method

17×40＝ 100－63＝ 3.2＋1.68＝ 2.8×0.4＝
14－7.4＝ 1.92÷0.04＝ 0.32×500＝ 0.65＋4.35＝
10－5.4＝ 4÷20＝ 3.5×200＝ 1.5－0.06＝
0.75÷15＝ 0.4×0.8＝ 4×0.25＝ 0.36＋1.54＝
1.01×99＝ 420÷35＝ 25×12＝ 135÷0.5＝
3/4 + 1/4 = 2 + 4/9 = 3 - 2/3 = 3/4 - 1/2=
1/6 + 1/2 -1/6 = 7.5-(2.5+3.8)= 7/8 + 3/8 =
3/10 +1/5 = 4/5 - 7/10 = 2 - 1/6 -1/3 =
0.51÷17＝ 32.8＋19＝ 5.2÷1.3＝ 1.6×0.4＝
4.9×0.7＝ 1÷5＝ 6÷12＝ 0.87－0.49＝
17×40＝ 100－63＝ 3.2＋1.68＝ 2.8×0.4＝
14－7.4＝ 1.92÷0.04＝ 0.32×500＝ 0.65＋4.35＝
10－5.4＝ 4÷20＝ 3.5×200＝ 1.5－0.06＝
0.75÷15＝ 0.4×0.8＝ 4×0.25＝ 0.36＋1.54＝
1.01×99＝ 420÷35＝ 25×12＝ 135÷0.5＝
3/4 + 1/4 = 2 + 4/9 = 3 - 2/3 = 3/4 - 1/2=
1/6 + 1/2 -1/6 = 7.5-(2.5+3.8)= 7/8 + 3/8 =
3/10 +1/5 = 4/5 - 7/10 = 2 - 1/6 -1/3 =
0.51÷17＝ 32.8＋19＝ 5.2÷1.3＝ 1.6×0.4＝
4.9×0.7＝ 1÷5＝ 6÷12＝ 0.87－0.49＝
17×40＝ 100－63＝ 3.2＋1.68＝ 2.8×0.4＝
14－7.4＝ 1.92÷0.04＝ 0.32×500＝ 0.65＋4.35＝
10－5.4＝ 4÷20＝ 3.5×200＝ 1.5－0.06＝
0.75÷15＝ 0.4×0.8＝ 4×0.25＝ 0.36＋1.54＝
1.01×99＝ 420÷35＝ 25×12＝ 135÷0.5＝
3/4 + 1/4 = 2 + 4/9 = 3 - 2/3 = 3/4 - 1/2=
1/6 + 1/2 -1/6 = 7.5-(2.5+3.8)= 7/8 + 3/8 =
3/10 +1/5 = 4/5 - 7/10 = 2 - 1/6 -1/3 =
0.51÷17＝ 32.8＋19＝ 5.2÷1.3＝ 1.6×0.4＝
4.9×0.7＝ 1÷5＝ 6÷12＝ 0.87－0.49＝
17×40＝ 100－63＝ 3.2＋1.68＝ 2.8×0.4＝
14－7.4＝ 1.92÷0.04＝ 0.32×500＝ 0.65＋4.35＝
10－5.4＝ 4÷20＝ 3.5×200＝ 1.5－0.06＝
0.75÷15＝ 0.4×0.8＝ 4×0.25＝ 0.36＋1.54＝
1.01×99＝ 420÷35＝ 25×12＝ 135÷0.5＝
3/4 + 1/4 = 2 + 4/9 = 3 - 2/3 = 3/4 - 1/2=
1/6 + 1/2 -1/6 = 7.5-(2.5+3.8)= 7/8 + 3/8 =
3/10 +1/5 = 4/5 - 7/10 = 2 - 1/6 -1/3 =
0.51÷17＝ 32.8＋19＝ 5.2÷1.3＝ 1.6×0.4＝
4.9×0.7＝ 1÷5＝ 6÷12＝ 0.87－0.49＝
17×40＝ 100－63＝ 3.2＋1.68＝ 2.8×0.4＝
14－7.4＝ 1.92÷0.04＝ 0.32×500＝ 0.65＋4.35＝
10－5.4＝ 4÷20＝ 3.5×200＝ 1.5－0.06＝
0.75÷15＝ 0.4×0.8＝ 4×0.25＝ 0.36＋1.54＝
1.01×99＝ 420÷35＝ 25×12＝ 135÷0.5＝
3/4 + 1/4 = 2 + 4/9 = 3 - 2/3 = 3/4 - 1/2=
1/6 + 1/2 -1/6 = 7.5-(2.5+3.8)= 7/8 + 3/8 =
3/10 +1/5 = 4/5 - 7/10 = 2 - 1/6 -1/3 =
0.51÷17＝ 32.8＋19＝ 5.2÷1.3＝ 1.6×0.4＝
4.9×0.7＝ 1÷5＝ 6÷12＝ 0.87－0.49＝
17×40＝ 100－63＝ 3.2＋1.68＝ 2.8×0.4＝
14－7.4＝ 1.92÷0.04＝ 0.32×500＝ 0.65＋4.35＝
10－5.4＝ 4÷20＝ 3.5×200＝ 1.5－0.06＝
0.75÷15＝ 0.4×0.8＝ 4×0.25＝ 0.36＋1.54＝
1.01×99＝ 420÷35＝ 25×12＝ 135÷0.5＝
3/4 + 1/4 = 2 + 4/9 = 3 - 2/3 = 3/4 - 1/2=
1/6 + 1/2 -1/6 = 7.5-(2.5+3.8)= 7/8 + 3/8 =
3/10 +1/5 = 4/5 - 7/10 = 2 - 1/6 -1/3 =

Two cylinders plus a ball is equal to four cones plus three cubes, four balls are equal to one cone plus a cylinder plus a cube, one cylinder connects a cone to two balls plus a cube. It is calculated that the mass of four cones is equal to the mass of several balls

The title should be:
"Two cylinders and one sphere are equal to four cones and three cubes, four spheres are equal to one cone and one cylinder and one cube, one cylinder and one cone is equal to two spheres and one cube. It is calculated that the mass of four cones is equal to the mass of several spheres
Let a cylinder be a, a sphere be B, a cone be C, and a cube be d
It is known that 2A + B = 4C + 3D,
4b=c+a+d,
a+c=2b+d,
Find out the relationship between C and B
If B-3 / 2 * C = 0, then the relationship between B and C is as follows:
b: C = 3:2, so the mass of four cones is the mass of six spheres

When the four little rabbits picked mushrooms to go back, they had a total of 72 mushrooms. Half of rabbit a couldn't eat and lost, and two of rabbit B lost. Rabbit C picked them up and put them in the basket. At this time, their numbers were exactly the same. While rabbit Ding picked some more on the way out of the forest to double the number of baskets. Finally, when they got out of the forest, their numbers were equal, How many mushrooms are there in each rabbit basket? How many after walking out of the forest?

Out of the forest, a: 32, B: 18, C: 14, D: 8, out of the forest, each rabbit 16
I won't explain the detailed process. The 4-element linear equation is very easy to calculate, but the title is very difficult to understand

When studying the relationship between the current in the conductor and the resistance of the conductor when the voltage of the conductor is constant, what is the function of the sliding rheostat?

First, set a voltage and measure the data
Then replace the resistor, and move the sliding rheostat to make the voltage representation equal to that just now, and record the data
and so on

Straight line y = kx-1 and hyperbola 4x ^ 2-9y ^ 2 = 36
(1) When k satisfies what value, the straight line and the right branch of hyperbola have two focal points
(2) When k satisfies what value, the left branch of a straight line and a hyperbola have two focal points

When a straight line passes through a fixed point (0, - 1), the asymptote of hyperbola is y = ± 2x / 3. First calculate the value of K when it is tangent, and substitute y = kx-1 into hyperbola to get: 4x & # 178; - 9 (kx-1) &# 178; = 36 (4-9k & # 178;) x & # 178; + 18kx-45 = 0 △ = 324k & # 178; + 180 (4-9k & # 178;) = 09k & # 178; + 5 (4-9k & # 178;) = 020 = 36

Derivation of fan ring area formula
S=π(r^2+R^2+Rl+rl)
The way I deduce is to subtract the small sector from the large sector
(let X be the generatrix of the small sector, R be the lower radius, R be the upper radius, and l be the generatrix of the sector ring)
That is s = 1 / 2 [2 π R (L + x)] - 1 / 2 (π Rx)
Large sector area - small sector area
Because of the existence of X, we can't simplify the formula of R, R and L
In addition, check Baidu know that others say "fan ring area formula can refer to trapezoidal area method"
That is to use the trapezoidal area formula
The arc belt with a small circle at the top and the arc belt with a big circle at the bottom is just one of the other two sides. "
Trapezoid formula is to divide trapezoid into two triangles and use triangle formula to find
But when you divide the fan ring into those two parts, I can't say what shape is obtained by using what formula?
And the arc length * bus?
(with Tucao, that answer to me is like the formula of polygon, which is to make complaints about the same length of the edges.
Make complaints about... Although it seems that after flattening the arc, it's almost the same?
The same as the derivation of the volume formula of the cone
V=1/3(S'+√S'S+S)h
I don't understand why the volume difference formula in the book is like this

(1) There is another condition for the derivation of fan ring: X / (L + x) = R / R  XR = (L + x) RX = LR / (R-R) the arc length of small sector is 2 π R, the arc length of large sector is 2 π R (it's the side expansion of the cone) ｜ s = (1 / 2) [2 π R (L + x)] - (1 / 2) (2 π Rx) = π [RL + (R-R) x] = π (RL + LR) = π L (R + R) (2) the cone

（x-2y）²+16y-8x+16

（x-2y）²+16y-8x+16
=（x-2y）²-8(x-2y)+16
=(x-2y-4)²

Conversion relationship between battery capacity ah and consumer w

Ah. Is the capacity of the battery. W is the power of the consumer

Party A and Party B conduct a go game. It is agreed that the winner of the first three games will win the game. At the end of the game, suppose that in one game, the probability of party a winning is 0,6, and the probability of Party B winning is 0,4. The results of each game are independent. It is known that in the first two games, Party A and Party B win one game each. (I) find out the probability of party a winning the game; (II) let ξ denote the period from the third game to the end of the game At the end of the game, calculate the number of games, and get the distribution column and mathematical expectation

AI denotes the event: a wins in the first inning, (I = 3,4,5) Bi denotes B wins in the second inning, j = 3,4 (1) B denotes the event: a wins this game, ∵ in the first two innings, a and B each win one inning, ∵ a needs to win two innings first in the later competitions, ∵ B = a3a4 + b3a4a5 + a3b4a5, because the results of each inning are independent, ∵ P (b) = P (a3a4) ）+P (b3a4a5) + P (a3b4a5) = 0.6 × 0.6 + 0.4 × 0.6 + 0.6 × 0.4 × 0.6 = 0.648 (2) ξ represents the number of games from the beginning of the third inning to the end of the game. From the above question, we can see that the possible values of ξ are 2 and 3. Due to the independence of each inning, we get the distribution column P (ξ = 2) = P (a3a4 + b3b4) = 0.52p (ξ = 3) = 1-p (ξ = 2) = 1-0.52 = 0.48  e ξ = 2 × 0.52 + 3 × 0.48 = 2.48

The formula of finding kg with known Joule

I think it's Einstein's mass energy equation q = MC2