810x811x812.x2011 divided by the maximum value of n power of 810 (the result is a positive integer) n

810x811x812.x2011 divided by the maximum value of n power of 810 (the result is a positive integer) n

[2011/3] - [809/3] = 670 - 269 = 401[670/3] - [269/3] = 223 - 89 = 134[223/3] - [89/3] = 77 - 29 = 48[77/3] - [29/3] = 25 - 9 = 16[25/3] - [9/3] = 8 - 3 = 5[8/3] - [3/3] = 2 -1 =1401+134+48+16+5+1=605...

When a truck runs 50 kilometers per hour, it goes 100 kilometers ahead. A car runs 90 kilometers per hour from the same place to the same direction as the truck. The car can catch up with the truck a few hours after starting
Equivalent!

Set up a car to catch up with the truck after X hours
The distance of car starting x hours 90x = the distance of truck = 100 + 50x
90x=100+50x
40x=100
X = 2.5 hours
A: the car can catch up with the truck after 2.5 hours

Triangle ABC, a = 80, B = 100, B = 30 degrees?

There's only one

The two passenger and freight cars leave each other at the same time. When they meet, the distance ratio of the two cars is 6:5. After meeting, the freight car is 12 kilometers faster than the passenger car per hour, and the passenger car is still moving at the original speed. As a result, the two cars arrive at each other's departure station at the same time. It is known that the freight car has been running for 10 hours. How many kilometers is the distance between the two places?

6 △ 5 = 1.2, 12 △ 1.2 × 1.2-1 × 1.2, = 12 △ 3625-1 × 1.2, = 12 △ 1125 × 1.2, ≈ 32.7 (km); distance between the two places: 32.7 × 10 = 327 (km); answer: the distance between the two places is 327 km

High school mathematics problem! The geometric meaning of absolute value?
Title: when x takes all real numbers, what is the maximum and minimum value of the formula / X-2 / - / x + 4 /?
Requirements: clear solution ideas! To process! Good words plus points!

The problem is well solved with the help of the number axis
/X-2 / - / x + 4 / means the difference between the distance from X to 2 and the distance from X to - 4
When x ≥ 2 / X-2 / - / x + 4 / = - 6
When - 4

Two cars set out from a certain place at the same time to transport a batch of goods to the construction site 165 kilometers away. Car a arrived 48 minutes earlier than car B. when car a arrived, car B was 24 kilometers away from the construction site. How many hours did car a spend on the whole journey?

48 minutes = 0.8 hours; 24 △ 0.8 = 30 (km / h); 165 △ 30-0.8, = 5.5-0.8, = 4.7 hours; a: it took 4.7 hours for car a to complete the whole journey

Square of 9 (a-b) + 30 (a-b) (a + b) + 25 (a + b)

The speed of train is 60% faster than that of car. How much slower is the speed of car than that of train?
What percentage of the speed of the car is slower than that of the train

If the car speed is 1, then the train speed is 1.6
Car speed is 1, slower than train speed = 1-1 / 1.6 = 3 / 8

Δ ABC, ∠ C = 90 °, CD ⊥ AB at D, make ∠ CDE = ∠ CDF = α, AC at F, BC at E
When the value of α is, the area of △ DEF is the largest and the largest area is calculated

First of all, because △ ABC is given, a, B, C, ab = C, BC = a, CA = B are all known quantities. Moreover, it is easy to know that CD = AB / C, ∠ DCB = ∠ a, ∠ ACD = ∠ B
In △ CDF, according to the sine theorem, we know that DF / SINB = CD / sin (α + b), that is, DF = CD * SINB / sin (α + b);
Similarly, in △ CDE, according to the sine theorem, there is de = CD * Sina / sin (α + a)
As we know, the area of △ DEF is s = de * DF * sin2 α / 2, so the area expression is
S=CD*SinA/Sin(α+A)*CD*SinB/Sin(α+B)*Sin2α/2
=(a^2b^2/2c^2)*SinASinBSin2α/Sin(α+A)Sin(α+B)
=(a^2b^2/2c^2)*SinASinB*2SinαCosα/(SinαCosA+CosαSinA)(SinαCosB+CosαSinB)
=(a^2b^2/c^2)*1/(CotA+Cotα)(TanαCotB+1)
=(a^2b^2/c^2)*1/(CotATanαCotB+CotB+CotA+Cotα)
Since cotatan α cotb + cot α ≥ 2 √ (cotacotb) = 2 (because a and B are complementary), so
(a^2b^2/c^2)*1/(CotATanαCotB+CotB+CotA+Cotα)
≤(a^2b^2/c^2)*1/(CotB+CotA+2).
The equal sign holds if and only if Tan α = cot α, that is, α = 45 °
Considering that COTA = B / A, cotb = A / B, the maximum area can also be written as
a^3b^3/c^2(a+b)^2.
So the conclusion is that when α = 45 ° the maximum area of triangle is a ^ 3B ^ 3 / C ^ 2 (a + b) ^ 2

A bus travels 40 kilometers per hour from place a to place B, and 60 kilometers per hour from place B to place a. the average speed of the bus is calculated

Suppose the distance between a and B is s, then the time from a to B is s / 40, the time from B to a is s / 60, the total time is s / 40 + S / 60, the total distance is 2S, so the average speed is 2S / (s / 40 + S / 60) = 48 km / h