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A 60 meter long fence is used to form a rectangular flower bed. If the length is 3 meters less than twice the width, what is the area of the rectangular flower bed?
60÷2=30m
Set the width to x [you can set the length, but it's not easy, so it's better to set the width]
x+（2x-3）=30
3x=30+3
x=11
Length: 30-11 = 19m
Area: 11 × 19 = 209 square meters
A: the area of this rectangular garden is 209 square meters
There is a fence of 24 meters long and a rectangular flower bed with a fence in the middle is enclosed by a wall (the maximum usable length of the wall is a is 10 meters)
As shown in the picture
A————————D
1 1 1
1 1 1
1 1 1
B————————C
(1) If the area of the flower bed is 24 square meters, calculate the width of the flower bed ab
(2) Can the area of the flower bed be 45 square meters? If so, the width of the flower bed should be ab
(3) Can the area of the flower bed be enclosed into 48 square meters? Yes, the width of the flower bed is ab. no, please explain the reason
The 24 square meters of the first question should be changed to 42 square meters
(1) Suppose: X is the length of the flower bed (perpendicular to the wall); the length of the fence in the middle is also X; y is the width of the flower bed (AB) (parallel to the wall). Then: xy = 42; 3x + y = 24
A rectangular flower bed is 8 meters long and 6 meters wide. One side of it is against the wall, and the other three sides are fenced. How many meters is the fence at least?
1.6 × 2 + 8 = 20m
2.8 × 2 + 6 = 22m
Simple operation of 13 + 17 + 21 + 25 + 29 + 33 + 37
(33+17)+(21+29)+(13+37)+25=50+50+50+25=175
(13+37)*7/2=175
In equilateral triangle ABC, BD = 1 / 3bC, CE = 1 / 3aC
It's AP vertical CP
P is the intersection of AD and be
The title is wrong. What point is p? It is not involved in the title
If plane vector AI satisfies | AI | = 1 (I = 1234) and vector AI * a (I + 1) = 0, then the maximum value of | a1 + A2 + a3 + A4 | is
RT
AI * a (I + 1) = 0, so AI ⊥ a (I + 1),
So A1 / / A3, A2 / / A4,
When A1 = A3, A2 = A4,
|A1 + A2 + a3 + A4 | has a maximum of 2 | a1 + A2|
And | a1 + A2 | &# 178; = (a1 + A2) &# 178; = A1 &# 178; + 2A1 * A2 + A2 &# 178; = 1 + 0 + 1 = 2
So | a1 + A2 + a3 + A4 | has a maximum value of 2 √ 2
We know that the area of a trapezoid is 9 square centimeters, its upper bottom is 4.5 centimeters, its lower bottom is 5.5 centimeters, how many centimeters is its height? Solve the equation
Set the height as H,
(4.5+5.5)*h/2=9, 5h=9, h=1.8
So the height is 1.8cm
Let a, B ∈ R, and a not equal to 2, f (x) = 1 + ax / 1 + 2x satisfy f (x) + F (- x) = 0
Let a, B ∈ R, and a not equal to 2, f (x) = LG (1 + ax / 1 + 2x) satisfy f (x) + F (- x) = 0 for the function defined in the interval (- B, b)
Find the value range of B
f(x)+f(-x)=0
1-a^2x^2=1-4x^2
(a^2-4)x^2=0
a=-2
The domain is (1-2x) / (1 + 2x) > 0
-1/2
In one corner of the warehouse, there is a pile of rice in a quarter cone shape. The arc length of the bottom surface is 3.14m, the height of the cone is 1m, and the weight of rice per cubic meter is 720kg
How many kilos of rice is this pile
Conical bottom area = (3.14 * 4 / π / 2) * (3.14 * 4 / π / 2) * π / 4 = 3.14 square meters;
Volume = cone = 3.14 * 1 / 3 = 1.0467 m3;
The weight of this pile of rice is 1.0467 * 720 = 753.6kg
Find the trajectory equation of the midpoint of the parallel chord with slope 2 in the ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1
y=2x+b
4x^2+9(2x+b)^2-36=0
40x^2+36bx+9b^22-36=0
x1+x2=-35b/40=-7b/8
x/2=-7b/16
(y1+y2)/2=y=(x1+x2)+b
=-7b/8+b=b/8
x/y=(-7/16)*8=-7/2
-7y=2x
2x+7y=0
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