Someone plays tennis from the origin o, and the flight path of the tennis ball is a parabola, which can be represented by the image of the quadratic function y = 4x-1 / 2x 2 To find the distance between the highest flying point B and the ground and the horizontal distance between the landing point a and the point o

Someone plays tennis from the origin o, and the flight path of the tennis ball is a parabola, which can be represented by the image of the quadratic function y = 4x-1 / 2x 2 To find the distance between the highest flying point B and the ground and the horizontal distance between the landing point a and the point o

y=4x-1/2x²
The axis of symmetry is x = 4
So when x = 4, y has a maximum value of 8
So the distance between the highest point B of tennis flight and the ground is 8
Let y=0 to get x=0 or 8
So the horizontal distance between a and O is 8

As shown in the figure, the quadratic function y = 1 4x2+(m The image of 4 + 1) x + m (m < 4) intersects with X axis at point a and B (1) Find the coordinates of points a and B (which can be expressed by algebraic expression containing the letter M); (2) If the graph of this quadratic function and the inverse scaling function y = 9 The image of X intersects at point C, and the cosine of ∠ BAC is 4 5. Find the analytic expression of this quadratic function

(1) When y = 0, 14x2 + (M4 + 1) x + M = 0, (1 point) x2 + (M + 4) x + 4m = 0, X1 = - 4, X2 = - M. (2 points) ∵ m ﹤ 4, ? a (- 4,0), B (- M, 0) (5 points) (2) pass through point C as CD ⊥ x-axis, and the vertical foot is D, cos ﹥ BAC = ADAC = 45. If ad = 4K, AC = 5K, then CD = 3K. (6 points) ? o

As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained

The transition point O is OC ⊥ AB to C, as shown in the following figure:  AOC = 12 ﹤ AOB = 60 °, AC = BC = 12ab, ﹤ in RT △ AOC, ﹤ a = 30 ° OC = 12oa = 10cm, AC = oa2 − oc2 = 202 − 102 = 103 (CM), ab = 2Ac = 203cm ﹥ the area of AOB = 12ab · OC = 12 × 203 × 10 = 1003 (cm2)

It is known that, as shown in the figure, BD is the bisector of ∠ ABC, ab = BC, point P is on BD, PM ⊥ ad, PN ⊥ CD, and the perpendicular feet are m and N respectively

It is proved that: in △ abd and △ CBD, ab = BC (known), ∠ abd = ∠ CBD (property of bisector), BD = BD (common side),  abd ≌ △ CBD (SAS), ∵ ADB = ∠ CDB (the corresponding angles of congruent triangles are equal); ∵ PM ⊥ ad, PN ⊥ CD, ∵ PMD = ∠ PND = 90 °; and ∵ PD = PD (...)

It is known that, as shown in the figure, BD is the bisector of ∠ ABC, ab = BC, point P is on BD, PM ⊥ ad, PN ⊥ CD, and the perpendicular feet are m and N respectively

It is proved that: in △ abd and △ CBD, ab = BC (known), ∠ abd = ∠ CBD (property of bisector), BD = BD (common side),  abd ≌ △ CBD (SAS), ∵ ADB = ∠ CDB (the corresponding angles of congruent triangles are equal); ∵ PM ⊥ ad, PN ⊥ CD, ∵ PMD = ∠ PND = 90 °; and ∵ PD = PD (...)

In the triangle ABC, D and F are on BC and AC respectively, ad and BF intersect at point E, and BD: DC = 3:2, AE = ed, then be: EF=____

In △ BCF, DG / BF = CD / BC,
But BD / DC = 3 / 2,  CD / BC = 2 / 5. ♫ dg / BF = 2 / 5
In △ ADG, AE = ed, ν EF = 1 / 2 * DG
Therefore, DG / BF = 2ef / BF = 2 / 5. ν EF / BF = 1 / 5
Therefore, be / EF = 4 / 1

It is known that in △ ABC, ∠ B = 90 ° AB = BC, DB = CE, M is the midpoint of the AC side. It is proved that △ DEM is an isosceles right triangle

Your question is not complete enough. I understand it. I can prove it
Join BM. Angle c = angle DBM, = 45 degrees. M is the midpoint of AC side, ab = BC. So BM = cm
So triangle DBM, congruent triangle MCE, angle DMB = angle EMC
DM = MC, so the triangle DEM is an isosceles triangle
Angle BMC = 90 degrees. Angle DMB = angle EMC, so angle DMC = 90 degrees
DEM is an isosceles right triangle

As shown in the figure, ∠ ABC = ∠ DBE = 90 °, DB = be, ab = BC, verification: ad = CE, ad ⊥ CE The second question: if △ DBE rotates around point B to the outside of △ ABC, and other conditions remain unchanged, is the conclusion in the first question true? Please prove it.

Because ∠ ABC = ∠ DBE = 90 °, so ∠ ABC - ∠ DBC = ∠ DBE - ∠ DBC, ∠ abd = ∠ CBE, and because DB = be, ab = BC, so triangle abd is all equal to triangle BCE, so ad = CE

In the RT triangle ABC, the angle ACB = 90 degrees, AC = BC, D is the midpoint of BC, CE ⊥ ad, and the perpendicular foot is e, In Rt triangle ABC, the angle ACB=90 degrees, AC=BC, D are the midpoint of side BC, CE is perpendicular to AD, E is perpendicular to E, BF is parallel to AC, the extension line crossing CE is parallel to point F, and DF is connected In triangle ACD, angle CAD = angle BCF AC = BC Angle ACD = angle CBF = 90 degrees Then: ACD of triangle is equal to CBF of triangle So: CD = BF CD = BD Then: BD = BF Then the triangle BDF is an isosceles right triangle And ab bisect angle DBF (angle DBA = angle ABF = 45 degrees) So AB bisects DF vertically however Why? In triangle ACD, angle CAD = angle BCF AC = BC Angle ACD = angle CBF = 90 degrees How to find the congruence of triangle ACD and triangle CFB?

Because BF is parallel to AC, angle ACD = angle CBF = 90 degrees
And angle CAD + angle ace = 90 degrees = angle BCF + angle ace, so there is: angle CAD = angle BCF
From the title, AC = BC

As shown in the figure, in RT △ ABC, the vertical bisector De of ∠ ACB = 90 ° intersects AC with E, and the extension of intersection BC is at F. if ∠ f = 30 ° de = 1, then the length of be is______ .

∵∠ACB=90°，FD⊥AB，
∴∠ACB=∠FDB=90°，
∵∠F=30°，
Ψ a = ∠ f = 30 ° (the remainder of the same angle is equal)
The vertical bisector De of AB intersects AC with E,
∴∠EBA=∠A=30°，
In the right angle △ DBE, be = 2DE = 2
So the answer is: 2