A village plans to excavate a 1500m long canal. The cross section of the canal is isosceles trapezoid. The depth of the canal is 0.8m, the width of the bottom is 1.2m, and the slope angle is 45 degrees. When the canal is actually excavated, 20 cubic meters more soil is excavated every day than the original plan. As a result, it is completed four days ahead of the original plan. How many cubic meters of soil is excavated every day?

The total amount of soil excavated should be: (1.2 + 2.8) × 0.8 △ 2 × 1500 = 2400 cubic meters. If the original plan is to excavate x cubic meters per day, then the actual excavation is (20 + x) cubic meters per day. According to the meaning of the question: 2400x − 2400x + 20 = 4, the solution is: x = 100. After testing, x = 100 is the root of the original equation

The cross section of a channel is trapezoidal, with a bottom width of a meter, a surface width of (a + 2b) meters and a channel depth of 0.5 meters (as shown in the figure)

S=1/2×[a+（a+2b）]×0.5a
=1/4×2（a+b）
=（a+b）/2

Dig a canal. It's a trapezoid. The upper bottom is 2.5 meters. The lower opening is 1.3 meters wide. The depth of the canal is 0.78 meters. What's the area of the canal

(2.5 + 1.3) x0.78/2 = 1.428 M2

It is known that a = a + 2. B = a ^ 2-A + 5. C = a ^ 2 + 5a-10, where a

B-A=a^2-2a+3=(a-1)^2+2>0 B>A
C-A=a^2+4a-12=(a+6)(a-2)
a0 C>A
When a = - 6, the above formula = 0, C = a
2> A > - 6

Given that a, B and C satisfy (A-1) &# 178; + 5 √ b-5a + | A-B + C + 1 | = 0, find the arithmetic square root of a + B + C

a-1=0
b-5a=0
a-b+c+1=0
∴a=1
b=5
c=3
∴√(a+b+c)=√(1+5+3)=3

Given a & # 178; - 5A + 1 = 13, find the value of 4A & # 178; - 20-1
Lost an a. Given a & # 178; - 5A + 1 = 13, find the value of 4A & # 178; - 20a-1

a²-5a+1=13
a²-5a=12
Double four on both sides
4a²-20a=48
So 4A & # 178; - 20a-1 = 47

If the quadratic power of a is known to be - 5A + 1 = 13, then the quadratic power of 4a is - 20a-1

∵a²-5a﹢1=13
∴a²-5a=12
∴4a²-20a=48
∴4a²-20a-1=48-1=47

How to prove this question? Proof: 0.09

The solution is as follows:
1/10²+1/11²+1/12²+… … +1/100²
1/10-1/101>0.09
So the original inequality is proved
The key to this problem is to make the denominator smaller or larger one by one, so that each item becomes two items after being staggered, and then offset each other, just like 1 / (9 * 10) = 1 / 9-1 / 10. It's not difficult to remember this

A = 0.09, B = 1 / (10) 2 + 1 / 11, C = 0.11

a＜b＜c
1/n^2＞1/n(n+1)=1/n-1/(n+1)
1/n^2＜1/n(n-1)=1/(n-1)-1/n
n=10、11、12...100
1/10-1/11

A = a plus 2, B = the square of a minus a plus 5, C = the square of a plus 5A minus 19, a is greater than 2, compare the size of a and B, a and C?

A is less than B
When a is greater than 2 and less than 3, a is greater than C
When a equals 3: a equals C
When a is greater than 3: A is less than C