A village plans to build a canal, its cross section is isosceles trapezoid, the bottom angle is 120 degrees, the sum of the two waist and the bottom is 4 meters. When the depth X of the canal is what, the cross-sectional area s is the largest, what is the maximum area?

A village plans to build a canal, its cross section is isosceles trapezoid, the bottom angle is 120 degrees, the sum of the two waist and the bottom is 4 meters. When the depth X of the canal is what, the cross-sectional area s is the largest, what is the maximum area?

It is known that if the waist length is 2x / √ 3, the lower sole is 4-2 * 2x / √ 3 = 4-4x / √ 3, and the upper sole is 4-2x / √ 3, then s = 1 / 2 * x (4-2x / √ 3 + 4-4x / √ 3) = - √ 3x ^ 2 + 4x,
When x = 2 √ 3 / 3, s is the largest, and the maximum is 4 √ 3 / 3
When the canal depth is 2 √ 3 / 3 m, the cross-sectional area s is the largest, and the largest area is 4 √ 3 / 3 m2

As shown in the figure, a village plans to build a canal, its cross section is isosceles trapezoid, the bottom angle is 120 degrees, the sum of two waist and bottom is 6m, what is the maximum area?
No formula

The upper base angle of isosceles trapezoid is equal to 60 degrees. If the length of the lower base is XM and the waist length is YM, then the lower base is x = 6-2y; the upper base is x + 2ycos60 degrees = x + y = 6-y; the depth (height) is ysin60 degrees = (√ 3 / 2) y, and the area is s = (6-2y + 6-y) (1 / 2) (√ 3 / 2) y = (√ 3 / 4) (12-3y) y
To get the maximum value of S, we need the maximum value of S1 = (12-3y) y, or S2 = 3s1 = (12-3y) * 3Y, because (12-3y) + 3Y = 12 is a fixed value, so when 12-3y = 3Y, that is, y = 2, S2 gets the maximum value, so S1 and s also get the maximum value
Smax=(√3/4)×（12-3×2）×2=（√3/4）×6×2=3√3m²

The area of a trapezoid is 70 square meters, the upper bottom is 5 meters, and the height is 10 meters. How many meters is the lower bottom?

Bottom = 70 × 2 △ 10-5 = 9m

Given X & # 178; + 4Y & # 178; - 6x + 4Y + 10 = 0, find the value of 1 / x + 1 / y

x²+4y²-6x+4y+10=
=(x^2-6x+9)+(4y^2+4y+1)
=(x-3)^2+(2y+1)^2=0
x=3 y=-1/2
1/x+1/y
=1/3-2
=-5/3

Given X & # 178; + Y & # 178; + 6x-4y + 13 = 0, find the 2014 times of the algebraic formula (x + y)

A:
x²+y²+6x-4y+13=0
(x²+6x+9)+(y²-4y+4)=0
(x+3)²+(y-2)²=0
So:
x+3=0
y-2=0
The solution is: x = - 3, y = 2
So: (x + y) ^ 2014 = (- 3 + 2) ^ 2014 = 1

X & # 178; + Y & # 178; + 6x-4y + 13 = 0, where x and y are rational numbers, then what is the x power of Y?

x²+y²+6x-4y+13=0
(x+3)^2+(y-2)^2=0
x=-3,y=2
y^x=2^(-3)=1/8

Given y = 3, find the value of 6x-3xy + 7y-3xy-4xy + 2Y

∵y/x=3∴y=3x
The original formula = (3x & # 178; - 12x & # 178; + 18x & # 178;) / (6x & # 178; - 3x * 3x + 7 * 9x & # 178;)
=9x²/(60x²)
=3/20

Given the quadratic power of X + the quadratic power of Y + 13 = 6x-4y, find the value of the quadratic power of (2x-y) - 2 (2x-y) (x + 2Y) + (x + 2Y)

Because x ^ 2 + y ^ 2 + 13 = 6x - 4Y
That is, x ^ 2 - 6x + 9 + y ^ 2 + 4Y + 4 = 0
That is, (x - 3) ^ 2 + (y + 2) ^ 2 = 0
Therefore, x = 3, y = - 2
So (2x - y) ^ 2 - 2 (2x - y) (x + 2Y) + (x + 2Y) ^ 2
= (2x-y - x -2y)^2
= (x -3y)^2
Substituting x = 3, y = - 2 into the above formula, we get
The original formula = [3-3 × (- 2)] ^ 2 = 81

Known a

[(a+b)/(a-b)]²
=(a+b)²/(a-b)²
=(a²+b²+2ab)/(a²+b²-2ab)
=(4ab+2ab)/(4ab-2ab)
=3
∵ a

Given | a-6 | + (B + 5) & # 178; = 0, find - B + A-2 / 3

Given | a-6 | + (B + 5) & # 178; = 0, find - B + A-2 / 3
According to the meaning of the title:
a-6=0;
a=6;
b+5=0;
b=-5;
∴-b+a-2/3
=5+6-2/3
=10+1/3
=10 and 1 / 3;
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