As shown in the figure, in rectangular ABCD, fold the triangle BCD along BD to the position of triangle BDE, be and ad intersect at o point, and connect AE (1) Proof: the quadrilateral ABDE is isosceles trapezoid; (2) If the angle DBC = 30 ° AB = 2, find the area of quad ABDE

As shown in the figure, in rectangular ABCD, fold the triangle BCD along BD to the position of triangle BDE, be and ad intersect at o point, and connect AE (1) Proof: the quadrilateral ABDE is isosceles trapezoid; (2) If the angle DBC = 30 ° AB = 2, find the area of quad ABDE

(1) BC is obtained from rectangle ABCD= AD.AB=CD=DE Because of folding, angle EBD = angle DBC, ad ∥ BC, angle ADB = angle CBD, so angle EBD = angle ADB, OB = OD, OA = OE, angle OAE = angle OEA, AE ∥ BD, and AB and de are not parallel, the quadrangle ABDE is an isosceles trapezoid

In the trapezoidal ABCD shown in the figure below, the lower bottom is twice the upper bottom, and E is the midpoint of ab. how many times is the area of trapezoidal ABCD larger than that of triangle BDE?

Because trapezoid and triangle are equal height
The area ratio of trapezoid ABCD to triangle BDE is the ratio of the sum of trapezoid's top and bottom to the length of triangle's bottom
That is (1 + 2): 1 = 3:1
The area of trapezoid ABCD is three times that of triangle BDE

In a tetrahedral ABCD, if e and F are the midpoint of edges AB and CD respectively, what is the angle between the antiplane line AF and CE?

AF, EC, BF and E were connected respectively to make AF parallel line and BF to g
Because eg is parallel to AF, the problem is transformed into finding the angle between eg and CE
The angle GEC is the included angle,
cosGEC=(EG^2+EC^2-GC^2)/2*EG*EC
Let the side length of tetrahedron be x, eg = √ 3 / 4 x; EC = √ 3 / 2 x; GC = √ 7 / 4 X,
So cosgec = 2 / 3
So the angle between AF and CE is arccos2 / 3

In a tetrahedron a-bcd, e is the midpoint of BC, and the cosine of the angle between AE and BD is obtained

Taking the midpoint F of CD, connecting EF and AF, we can get that E and F in ∵ BCD are the midpoint of BC and CD respectively, ∵ EF ∥ BD, EF = 12bd. Therefore, ∵ AEF (or its complementary angle) is the angle formed by the out of plane straight line AE and BD. let the edge length of the regular tetrahedron be a, we can get AF = AE = 32a, EF = 12a from the title. In ∵ AEF, according to the cosine theorem, we can get cos ∵ AEF = ef2 + EA2 − af22ef · EA = 14a2 + 34a2 − 34a22 × 12a × 32A = 36 The cosine value of the angle between AE and BD is 36

Find the inverse function of y = x & # 178; - 4x + 5 (x ≤ 1),

y-1=(x-2)^2
x-2

Inverse function of y = 2 / 1 (e ^ x-e ^ - x)?

Let t = e ^ x, then t > 0
y=1/2*(t+1/t)
T = y + (y ^ 2 + 1) ^ (1 / 2) because t > 0
X = log (y + (y ^ 2 + 1) ^ (1 / 2)) y from negative infinity to positive infinity

How to find the inverse function of y = 1 / 2 (ex + E-X)

y=[e^x+e^(-x)]/2
2y = e^x + e^(-x)
2y e^x = (e^x)^2 + 1
(e^x)^2 - 2y e^x + 1 = 0
e^x = [2y ±√(4y^2 -4)]/2 = y ±√(y^2 -1)
among
y -√(y^2 -1)
= 1/[y +√(y^2 -1)]
< 1/y
meanwhile
y=[e^x+e^(-x)]/2
≥2 √[e^x * e^(-x)] /2
= 1
When x = 0, y = 1
And x > 0
therefore
y > 1
1/y < 1
y -√(y^2 -1) < 1
And x > 0, e ^ x > 1
Therefore, e ^ x = y - √ (y ^ 2 - 1) is omitted
e^x = y +√(y^2 -1)
x = ln [y +√(y^2 -1)]
Interchange X and Y symbols
y = ln [x -√(x^2 -1)]
Where x > 1

What is the inverse function of y = (1 / 2) (e ^ X-1 / e ^ x)?

y=(1/2)(e^x-1/e^x)
e^(2x)-2ye^x-1=0
e^x=y+(4y^2+4)^0.5=y+(y^2+1)^0.5
x=ln[y+(y^2+1)^0.5]
So the inverse function is
y=ln[x+(x^2+1)^0.5]

The inverse function of the function y = E2x (x ∈ R) is ()
A. y=2lnx（x＞0）B. y=ln（2x）（x＞0）C. y=12lnx（x＞0）D. y=12ln（2x）（x＞0）

From y = E2x, we can get 2x = LNY, that is & nbsp; X = 12lny. By exchanging X and y, we can get y = 12lnx (x ＞ 0). The inverse function of y = E2x (x ∈ R) is y = 12lnx (x ＞ 0)

Finding the inverse function of y = ex + 1 / EX-1
Ex is the x power of E

Let a = e ^ x
y=(a+1)/(a-1)
ay-y=a+1
ay-a=y+1
a=e^x=(y+1)/(y-1)
So the inverse function is e ^ y = (x + 1) / (x-1)
J is y = ln [(x + 1) / (x-1)]