In the RT triangle ABC, the angle BAC = 90, ad ⊥ BC at point D, e is the midpoint of AD, the extension line connecting ed and intersecting AB at point F, AB / AC = DF / AF is proved E is the midpoint of AC

In the RT triangle ABC, the angle BAC = 90, ad ⊥ BC at point D, e is the midpoint of AD, the extension line connecting ed and intersecting AB at point F, AB / AC = DF / AF is proved E is the midpoint of AC

It is proved that de = AE = (?) 189; AC ﹥ DAE = ﹥ ade. CAF = ﹥ ADB = 90 ﹥ DAE + ﹥ CAF = ﹥ ade. + ﹥ ADB is ﹥ BDF = ﹥ DAF in triangle BDF and triangle DAF, ﹥ BDF = ﹥ DAF, ﹥ f is the common angle ﹥ triangle BDF ﹥ triangle DAF ﹥ DF
I don't understand that? E is the midpoint of AD and needs to be connected to ed. A is the point on AB and the point on the extended line of ED. AB and ED can also intersect at f?
Can there be two intersections when two non coincident lines intersect???
There is something wrong with the title
If the tangent slope of the curve y = f (x) at the point (x0, f (x0)) is k, then Lim ∨ x → 0 f (x0 + △ x) - F
If the tangent slope of the curve y = f (x) at the point (x0, f (x0)) is k, then Lim ∨ x → 0 f (x0 + △ x) - f (x0) / △ X=_____ Seeking process
Y '= K (x) = (2 / x) + 2x ≥ 2 √ (2 / x) × 2x = 4 (x > 0)
From the meaning of the title, we know that K ≤ 4
So k = 4, x = 1, y = 1
Tangent equation: Y-1 = 4 (x-1)
That is y = 4x-3
Let {& nbsp; an} be equal ratio sequence and {BN} be equal difference sequence, and B1 = 0, CN = an + BN, if {& nbsp; CN} is 1, 1, 2 To find the sum of the first 10 terms of the sequence {& nbsp; CN}
According to the meaning: C1 = a1 + B1 = 1, ∵ B1 = 0, ∵ A1 = 1, let & nbsp; BN = B1 + (n-1) d = (n-1) d (n ∈ n *), an = A1 · QN-1 = QN-1, (n ∈ n *) ∵ C2 = A2 + B2, C3 = A3 + B3, ∵ 1 = D + Q, 2 = 2D + Q2, the solution is: q = 0, d = 1, or q = 2, d = - 1 ∵ Q ≠ 0, ∵ q = 2, d = - 1
Solve linear equations X1 + x2 + X3 = 6,2x1 + 3x2 + 3x3 = 15,2x1 + 2x2 + 3x3 = 13
Ask for detailed explanation
Multiply the two sides of Formula 1 by 2 to: 2x1 + 2x2 + 2x3 = 12, and subtract formula 3 from this formula to get: X3 = 1
If both sides of Formula 1 multiply by 3, it becomes: 3x1 + 3x2 + 3x3 = 18. If this formula is subtracted from formula 2, it gets: X1 = 3, then x2 = 2
In the triangle ABC, the angle BAC is 90 degrees, ad is perpendicular to BC at D, e is the midpoint of AC, connects ed and extends the extension line of intersection AB at F, and proves AB times AF = AC times DF
Certificate:
Because ad, BC, e is the midpoint of AC
So CE = ed
So angle c = angle CDE
Because the angle BAC = 90 degrees, ad, BC
So angle bad = angle c
So angle FDB = angle fad
So triangle FDB is similar to triangle fad
So DF / AF = BD / ad
Because angle bad = angle c, angle ADB = angle BAC = 90 degrees
So triangle abd is similar to triangle CBA
So AB / AC = BD / ad
So DF / AF = AB / AC
So AB times AF = AC times DF
Given that the tangent slope at any point (x0, f (x0)) of the function f (x), (x belongs to R) is k = (x0-2) * (x0-3) ^ 2, then the monotone decreasing interval of the function is
It is easy to know that the derivative function f '(x) = (X-2) (x-3) ^ 2. = = = > when x, the decreasing interval of the function is (- ∞, 2)
Let {an} be an equal ratio sequence, {BN} be an equal difference sequence, and B1 = 0. The first three terms of the sequence {CN} are 1,1,2 in turn, and CN = an + BN
Then the sum of the first ten terms of {CN} is
A.467
B.978
C.988
D.968
B 978
an=2^(n-1)
bn=1-n
C10=A10+B10=978
Solving linear equations: X1 + 2x2 + X3 = 8, 2x1 + 3x2 + X3 = 11, X1 + 3x2 + 3x3 = 16
x+2y+z=8 (1)
2x+3y+z=11 (2)
x+3y+3z=16 (3)
(3) (1), y + 2Z = 8
2 * (1) - (2), y + Z = 5
Two equations subtract, z = 3, so y = 2, x = 1
The solutions of the equations are X1 = 1, X2 = 2, X3 = 3
In the triangle ABC, the angle BAC = 90 ° ad is perpendicular to BC and at point D, and E is the midpoint of AC. it is proved that ab × AF = AC × DF
prove:
Because ad is perpendicular to BC and to point D
Therefore, BDC = 90 degree
Because in the triangle ABC, ∠ BAC = 90 degree
Therefore, BDC = BAC = 90 degree
Because ∠ BDC = ∠ BAC, ∠ bad = ∠ bad
So triangle BDA is similar to BAC
So ∠ bad = ∠ C, AB / AC = BD / ad
Because point E is the midpoint of the hypotenuse AC
So de = CE = AE
So ∠ C = ∠ EDC
Because ∠ BDF = ∠ EDC
Therefore, C = BDF
Because ∠ C = ∠ bad
Therefore, BDF = bad
Because ∠ f = f, ∠ BDF = bad
So triangle FBD is similar to FDA
So BD / ad = DF / AF
Because BD / ad = AB / AC
So AB / AC = DF / AF
So ab × AF = AC × DF
Where is point F
Do you have a picture?: Yes, but I can't pass it
Then let f (T) + 0 (T) + 0 (T) + 0 (T) + 0 (T) + 0 (T) + 0 (T) + 0 (T) + 0 (T) + 0 (T) + 0 (T) + 0 (T) + 0 (t) + 0 (T) + 0 (T) + 0 (T) + 0 (t=
Let a function be differentiable at x0, then LIM (t → 0) [f (XO + T) + F (x0-3t)] / T=
plus
f(x0+t) = f(x0) + t f'(x0) + o1(t)
f(x0-3t) = f(x0)- 3t f'(x0) + o2(t)
F (x0 + T) + F (x0-3t) = 2F (x0) - 2T f '(x0) + O1 (T) + O2 (T)
Divide both sides by T to get [f (x0 + T) + F (x0-3t)] / T = 2F (x0) / T - 2 F '(x0) + [O1 (T) + O2 (T)] / T
When f (x0) is not 0, the value is infinite; when f (x0) is 0, the value is - 2 F '(x0)
The above o 1 (T) and O 2 (T) are higher order infinitesimals of T
In the middle is the minus sign
If it's a minus sign, it's 4f '(x0)
If it is a plus sign, the limit only exists when f (x0) is equal to 0. The limit is - 2F '(x0) by using the law of lobita