In rectangular paper ABCD, ab = 3cm, BC = 4cm. Now fold and flatten the paper so that a and C coincide. If the crease is EF, the area of the overlapping part △ AEF is equal to______ ．

Let AE = x, from folding, EC = x, be = 4-x, in RT △ Abe, AB2 + be2 = AE2, that is 32 + (4-x) 2 = X2, the solution is: x = 258, from folding, we can know ∠ AEF = ∠ CEF, ∫ ad ‖ BC, ∫ CEF = ∠ AFE, ∫ AEF = ∠ AFE, that is AE = AF = 258, ∫ s △ AEF = 12 × AF × AB = 12 × 258 × 3 = 7516

We know the function f (x) = - 2x + m, where m is a constant. When f (x) is an odd function, we find the value of the real number m?

If f (x) is an odd function, f (- x) = - f (x)
－2(-x)＋m=-(－2x＋m)
m=0

1995-1+2-3+4-5+… +1948-1949=______ ．

1995-1+2-3+4-5+… +1948-1949=1995-（1+1949+3+1947+… +975）+2+1948+4+1946+… +974 + 976 = 1995-975 × 975 + 975 × 974 = 1995-975 × (975-974) = 1995-975 = 1020

Factoring factor X ^ 4-3x ^ 2 + 2 in real number range

Original = (X & # 178; - 1) (X & # 178; - 2)
=(x + 1) (x-1) (X-2) (x + 2)

In cube, ABCD_ In a1b1c1d1, AC vertical BD1?

Because dd1 ⊥ face ABCD
And AC is on ABCD
So dd1 ⊥ AC
Because of AC ⊥ BD
So AC ⊥ surface bdd1
BD1 is on bdd1
So AC ⊥ BD1

-The third power of B is multiplied by the second power of B
How much is the answer

-The fifth power of B

The length of the oblique line PA passing through a point p outside the plane is (2 √ 3) / 3 times that of the vertical line Pb passing through this point (a, B ∈ R)
The length of the oblique line PA passing through a point p outside the plane is (2 √ 3) / 3 times that of the vertical line Pb passing through this point
（A、B∈R）
Find the angle between oblique line PA and plane
I'm so anxious

In RT triangle, if the ratio of three sides is 1: root 3: 2
The triangle is 30 degrees, 60 degrees and 90 degrees
So the angle between the diagonal PA and the plane is 60 degrees

Factorization: X4 + 6x + X + 12
How did you come up with this solution

1. Because x ^ 4 = x ^ 2 * x ^ 2; 12 = 4 * 3
Let's suppose that the original formula can be changed into: [as follows]
X^4+6X^2+X+12
=(x²+ax+4)(x²+bx+3)
=X^4+(a+b)X^3+(ab+7)X^2+(3a+4b)X+12
From this we can get the following conclusions
a+b=0
ab+7=6
3a+4b=1
When: a = - 1, B = 1, all three expressions hold
X^4+6X^2+X+12
=(x²-x+4)(X^2+X+3）
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The base of a parallelogram is 15 cm, the height is 6 cm, and a diagonal is 10 cm

First, draw a graph on the paper and mark all the known conditions, because the height is 6 and the diagonal is 10. You will see a right triangle. According to the square of the opposite side, it is equal to the square of the hypotenuse_ The square of the other pair of edges can be 8, because the bottom edge is 15, which can be 15_ 8 = 7, which forms a right triangle, find out the hypotenuse is the root 85, the four sides are out of the root 85, with 85, 15, 15, etc. and 30 + 2 with 85

In rectangular paper ABCD, ab = 3cm, BC = 4cm. Now fold and flatten the paper so that a and C coincide. If the crease is EF, the area of the overlapping part △ AEF is equal to______ ．