Plus 2x-1, the polynomial that equals 3x2-x-3 is () A. 3x2+x-4B. 3x2-3x-4C. 3x2-3x-2D. 3x2+x+2

Plus 2x-1, the polynomial that equals 3x2-x-3 is () A. 3x2+x-4B. 3x2-3x-4C. 3x2-3x-2D. 3x2+x+2

According to the meaning of the title, (3x2-x-3) - (2x-1) = 3x2-x-3-2x-1 = 3x2-3x-2

If f (x) belongs to R + for any x, f (x1 * x2) = f (x1) + F (x2), f (8) = 3, then f (radical 2)=

The function f (x) has f (x1 * x2) = f (x1) + F (x2) for any x belonging to R,
If f (8) = 3, then f (√ 2) =?
Because the function f (x) belongs to R for any x, f (x1 * x2) = f (x1) + F (x2)
So f (x) = loga (x)
Because loga (8) = 3, so a = 2, f (x) = log2 (x)
So f (2) = log2 (√ 2) = 1 / 2
If you don't understand, you are welcome to ask,

For any x1, X2 ∈ R, the function f (x) always has f (x1 + x2) = f (x1) f (x2). If f (1) = 2 under the root, then f (6) =?

f（6）=f（1+5）=f（1）f（5）
f（5）=f（1+4）=f（1）f（4）
f（4）=f（1+3）=f（1）f（3）
f（3）=f（1+2）=f（1）f（2）
f（2）=f（1+1）=f（1）f（1）
So f (6) = the sixth power of F (1) = 8

If we know that f (x) belongs to R for any x, we always have f (x1 times x2) = f (x1) + F (x2), if we know that f (8) = 3, we can find f (root 2), f (root 2), f (root 2)

If you choose to fill in the blanks, you can know that it is a logarithm algorithm according to f (x1x2) = f (x1) + F (x2), Let f (x) = log (a, x), and a be the base, then log (a, 8) = 3, a ^ 3 = 8, so a = 2, so f (x) = log (2, sqrt (2)) = 0.5; but you can't write the big problem like this: F (8) = f (4 * 2) = f (4) + F (2) = f (2 * 2) + F (2) = 3f (2) = 3, Then f (2) = 1, f (2) = f (root 2 * root 2) = f (root 2) + F (root 2) = 2F (root 2) = 1, so f (root 2) = 0.5, the calculation method of choosing to fill in the blanks can not be used for big problems, because there are other operations that follow this calculation method

Given that the diameter of circle O is ab = 10, the distance between point C and ab of chord CD is 3, and the distance between point D and ab is 4, then the distance between center O and chord CD is 3=

The perpendicular foot of point D on AB is e
The perpendicular foot of point C on AB is f
The perpendicular foot of point o on CD is g
CF=3 ,OF=4
ED=4,OE=3
So △ ode ≌ △ COF
∠COF=∠ODE
∠COF+∠EOD=90°
∠COD=90°
The △ OCD is an isosceles right triangle with 5 sides
OG=5*(√2)/2

Given that the radius of ⊙ o is 5cm, the chord ab ∥ CD, ab = 6cm, CD = 8cm, then the distance between AB and CD is ()
A. 1 & nbsp; CMB. 7 & nbsp; CMC. 1 & nbsp; cm or 7 & nbsp; CMD

It can be divided into two cases: ① when AB and CD are on the same side of O, as shown in Figure 1, through O, OE ⊥ AB is on e, intersecting CD is on F, connecting OA, OC, ∵ ab ∥ CD, ∥ of ⊥ CD, ∥ AE = 12ab = 3cm, CF = 12CD = 4cm, in RT △ OAE, OE = oa2 − AE2 = 52 − 32 = 4 (CM) is obtained by Pythagorean theorem, of = 3cm, EF = 4cm-3cm = 1cm; ② when AB and CD are on both sides of O, as shown in Figure 2, the same method If OE = 4cm, of = 3cm, then EF = 4cm + 3cm = 7cm; that is, the distance between AB and CD is 1cm or 7cm, so C

As shown in the figure, in the circle O, AB and CD are two equal chords, and ab ⊥ CD, the perpendicular foot is the point P, through the center O, make the perpendicular lines OE and of to AB and CD respectively
(1) Join OP, known OP = 3, root 2, OA = 5, find the length of ab

∵OP=3√2
∴OE=3
∴AE=√（5²－3²）=4
∴AB=8

As shown in the figure, in ⊙ o, diameter AB = 10, chord CD ⊥ AB, perpendicular foot is point E, if OE = 3, then CD=______ ．

Connecting OC, ∵ diameter AB = 10, ∵ OC = 12ab = 5, ∵ CD ⊥ AB, OE = 3, ∵ CD = 2ce, in RT △ OCE, CE2 + oe2 = oc2, that is CE2 + 32 = 52, the solution is CE = 4, ∵ CD = 2ce = 2 × 4 = 8

As shown in the figure, in the circle O, the chord AC ⊥ BD, and OE ⊥ CD is in E. if the length of AB is 10, then the length of OE is ()

As shown in the figure, in the circle O, the chord AC ⊥ BD, and OE ⊥ CD is in E. if the length of AB is 10, then the length of OE is (5)
As shown in the figure, make the diameter CF and connect DF,
Then ∠ ADF = 90 °,
∴∠1+∠2=90°,
And ∵ 3 + 4 = 90 °, 2 = 4,
∴∠1=∠3,
Ψ arc DF = arc AB,
∴DF=AB,
⊙ OE ⊥ CD in E,
∴CE=DE,
And ∵ co = fo,
∴OE=1/2DF,
∴OE=1/2AB=5

As shown in the figure, CD is the diameter of circle O, CD = 10, chord AB is vertical, CD, e is perpendicular, ab = 8, then OE =?

OE^2=OB^2-BE^2
OE=3