Who can help The side length of the square ABCD is 2cm. P is a point on the CD. The extension line connecting AP and BC intersects point E. when point P moves on the side CD, the area of the triangle Abe changes 1. Let PD = xcm (0 < x ≤ 2), find the functional relationship between the area y and X of the triangle Abe 2. According to the functional relationship in 1, determine where the point P is a triangle and the area is 400cm ^ 2 The square ABCD is marked counterclockwise from the upper left corner, which seems to affect the topic

Who can help The side length of the square ABCD is 2cm. P is a point on the CD. The extension line connecting AP and BC intersects point E. when point P moves on the side CD, the area of the triangle Abe changes 1. Let PD = xcm (0 < x ≤ 2), find the functional relationship between the area y and X of the triangle Abe 2. According to the functional relationship in 1, determine where the point P is a triangle and the area is 400cm ^ 2 The square ABCD is marked counterclockwise from the upper left corner, which seems to affect the topic

(1) The equation can be obtained by using similar triangles: y = 4 / X. (2) substitute 400 into x = 0.01 of the equation

Known a b+c＝b a+c＝c A + B = k, then the straight line y = KX + 2K must pass () A. Quadrants 1 and 2 B. Quadrants 2 and 3 C. Quadrants 3 and 4 D. Quadrants 1 and 4

Discussion by case:
When a + B + C ≠ 0, according to the proportional property of proportion, k = a + B + C
2(a+b+c)＝1
2. At this time, the straight line is y = 1
2X + 1, the straight line must pass through quadrants 1, 2 and 3
When a + B + C = 0, that is, a + B = - C, then k = - 1. At this time, the straight line is y = - X-2, that is, the straight line must pass through quadrants 2, 3 and 4
Combining the two cases, the straight line must pass through quadrants 2 and 3
Therefore, B

Using sequence limit to prove Lim 3N + 1 / 4N-1 = 3 / 4 Solve the problem and get 7 / 4n-11 / 4 (7/ ε+ 1) Why n = [1 / 4 (7/ ε+ 1）]+1 Why + 1?

If you want to reduce this problem directly to 7 / (16N - 4), n tends to infinity, you can prove it
n>1/4(7/ ε+ 1)
N=[1/4（7/ ε+ 1）]+1
It's rounding. Take an integer larger than it

Using the definition of sequence limit, it is proved that LIM (3N + 1) / (2n + 1) = 3 / 2 when n tends to positive infinity

Divide the fraction up and down by n to get LIM (3 + 1 / N) / (2 + 1 / N). Because when n tends to positive infinity, 1 / N = 0, so the equation = 3 / 2

Find the limit of the following sequence: LIM (2 + 3 ^ n) / (1 + 3 ^ (n + 1))

lim (2+3^n)/(1+3^(n+1))
Divide by 3 ^ n
=lim [(2/3^n)+1]/[(1/3^n)+3]
=(0+1)/(0+3)
=1/3
If you don't understand, you're welcome to ask

Find the limit LIM (x - > 0) [√ (1 + TaNx) -√ (1 + SiNx)] / [x √ (1 + sin ² x)-x]

lim(x->0) [√(1+tanx)-√(1+sinx)]/[x√(1+sin ² x)-x]
=lim [√(1+tanx)-√(1+sinx)]*[√(1+tanx)+√(1+sinx)] / [x√(1+sin ² x)-x]*[√(1+tanx)+√(1+sinx)]
=lim [(1+tanx)-(1+sinx)]*[√(1+sin ² x)+1] / x*[sin ² x]*[√(1+tanx)+√(1+sinx)]
=lim [tanx-sinx]*[√(1+sin ² x)+1] / x*[sin ² x]*[√(1+tanx)+√(1+sinx)]
=lim [tanx-sinx] / x*[sin ² x] * lim [√(1+sin ² x)+1] / [√(1+tanx)+√(1+sinx)]
=lim [tanx-sinx] / x*[sin ² x]
=lim [1/cosx - 1] / x*sinx
According to the equivalent infinitesimal
=lim x^2 / 2x^2
=1/2
If you don't understand, you're welcome to ask