Simple calculation: minus three and one-third times zero point 73 minus three and one-third times zero point 27; minus three fifths times three fifths plus nine fifths Minus three and one-third times zero point 73 minus three and one-third times zero point 27; Minus three fifths times three fifths plus nine fifths times minus twenty-four

Simple calculation: minus three and one-third times zero point 73 minus three and one-third times zero point 27; minus three fifths times three fifths plus nine fifths Minus three and one-third times zero point 73 minus three and one-third times zero point 27; Minus three fifths times three fifths plus nine fifths times minus twenty-four

Minus three and one-third times zero point 73 minus three and one-third times zero point 27
=Minus three and one third times (0.73 + 0.27)
=Minus three and one-third times one
=Negative three and one third
Minus three fifths times three fifths plus nine fifths times minus twenty-four
=Nine out of twenty-five plus nine out of twenty-five times twenty-four
=9 / 25 times (minus 1 + minus 24)
=Nine out of twenty-five times minus twenty-five
=Minus nine

Simple calculation of 1 × 2:1998 + 2 × 3:1998 + 3 × 4:1998 + 4 times 5:1998 + 5 × 6:1998

1. This problem needs to be solved by dividing the score into two parts, that is to say, dividing the score into the form of subtracting two parts, so as to make the positive and negative offsets
2. Because:
1/(1×2)=1－1/2；
1/(2×3)=1/2－1/3；
1/(3×4)=1/3－1/4；
1/（4×5）=1/4－1/5；
1/（5×6)=1/5－1/6.
1998/(1×2)+1998/(2×3)+1998/(3×4)+1998/（4×5）+1998/（5×6）
=1998×【1/(1×2)+1/(2×3)+1/(3×4)+1/（4×5）+1/（5×6)】
= 1998×（1－1/2＋1/2－1/3＋1/3－1/4＋1/4－1/5＋1/5－1/6）
=1998×（1－1/6）
=1998×5/6
=1665

45.16. For the function f (x) = {SiNx, SiNx ≥ cosx, cosx, SiNx < cosx, the following four propositions are given:
45.16. For the function f (x) = {SiNx, SiNx ≥ cosx, cosx, SiNx < cosx, the following four propositions are given
① The image of this function is symmetric with respect to x = 2K π + π / 4 (k belongs to Z)
② If and only if x = k π + π / 2 (k belongs to Z), the maximum value of this function is 1;
③ The function is a periodic function with π as the minimum positive period;
④ If and only if 2K π + π < x < 2K π + 3 π / 2 (k belongs to Z) - then √ 2 / 2 ≤ f (x) < 0
The correct serial number in the above proposition is (1) and (4)

This problem requires that f (x) represents the smaller part of the two functions
So you first draw the SiNx and cosx functions and then take the larger part
You can get a picture of the first floor by subtracting the others
Then analyze according to the graph

The power of 3333 to 5555 + the power of 5555 to 3333 divided by 7 is the remainder

3333^5555+5555^3333
=(3332+1)^5555+(5551+4)^3333,
Because 3332 and 5551 are multiples of 7,
The remainder of the preceding term divided by 7 is 1 according to the binomial theorem,
According to the binomial theorem, the remainder of the following term divided by 7 is 4 ^ 3333,
4^3333=64^1111=（7*9+1）^1111
Remainder 2

The turntable of the electric energy meter turns 600 turns in 5 minutes. What is the electric power in the home circuit + the turntable of the electric energy meter turns 600 turns in 5 minutes

Suppose the watt hour meter is "1 kwh / 3000 rpm" (there are also 2500 rpm, the algorithm is similar)
So, five minutes of hard work: 600 / 3000 = 0.2 kwh
That is, 0.2 kwh = 0.2 * 1000 * 3600 = 720000 joules
The electric power in this home circuit is 720000 joules / (5 * 60 seconds) = 2400 watts

On the existence theorem of zero point
Theorem (zero point theorem) Let f (x) be continuous on a closed interval [a, b], and f (a) and f (b) have different signs (i.e. f (a) × f (b))

F (a) and f (b) tell you that it's not equal to 0. There's no need to use a closed interval

Know the pump head and flow to get the calculation formula of pump motor power
I want to calculate the formula

9.81 * flow * head / 3600 / efficiency

C of X (a-b), a of Y (B-C), B of Z (C-A)

C: C / [x (a-b)] = C · y (B-C) · Z (C-A) / [x (a-b)]
A: A / [y (B-C)] = a · x (a-b) · Z (C-A) / [y (B-C)]
B: B / [Z (C-A)] = a · x (a-b) · CY (B-C) / [Z (C-A)]

The maximum allowable voltage at both ends of each festival lantern is 3V. If you want to connect them to the home circuit, you should at least string several such lamps
220 V / 3 = 73.333 why is the answer 74

You can't take 0.333333 bulbs, you can only take an integer, right? If you take 73 according to the general rounding principle, then the voltage of no bulb will be greater than 3V, exceeding the maximum allowable value of the bulb itself, so you can't take 73, but take 74, which is the most appropriate value to ensure the normal operation of the color lamp This class cannot follow rounding

Given that the calculation result of (x + 1) (X & # 178; + MX + n) does not contain X & # 178; term and X term, find m, n

The result is X & # 179; + (M + 1) x & # 178; + (M + n) x + n. since there are no X & # 178; and X terms, M + 1 = 0; m + n = 0. So m = - 1, n = 1