Finding the maximum value of the function y = x ^ 2-3x + 2 in the interval [1 / 2,2]

Finding the maximum value of the function y = x ^ 2-3x + 2 in the interval [1 / 2,2]

3/4

Function y = 3x (2-x) (0

y=3x(2-x)
=-3x²+6x
=-3(x-1)²+3
When x = 1,
The maximum value of Y is y = 3

Let a, B, C be the side lengths of triangle ABC, and for any real number x, f (x) = b2x2 + (B2 + c2-a2) x + C2 have ()
A. f（x）=0B. f（x）＞0C. f（x）≥0D. f（x）＜0

In △ ABC, according to the cosine theorem A2 = B2 + c2-2bccosa, ｜ B2 + c2-a2 = 2bccosa, the function can be reduced to: F (x) = b2x2 + (2bccosa) x + C2, ∵ B2 > 0 △ = 4b2c2cos2a − 4b2c2 = 4b2c2 (cos2a − 1) ＜ 0, and the image of the function y = f (x) is a parabola with an opening upward, and has no common point with the X axis

In the triangle ABC, points D and E are the middle points of edges BC and ab respectively. AD and CE intersect at point O, ab = 3, AC = 4, BC = 5. Find the length of OE

AB^2+AC^2=BC^2
Δ ABC is a right triangle, ∠ a = 90 & # 186;
CE^2=AC^2+AE2=4^2+(3/2)^2=73/4
CE=√73/2
OE=1/3*CE
=√73/6

Let a = (2,1), ab = 10., {a + B} = 5 times root 2?
How did you get 5 + 2 * 10 + B ^ 2

|a+b|^2=a^2+2ab+b^2
50=5+2*10+b^2
b^2=25
|b|=5

As shown in the figure, in △ ABC, ∠ a = 60 °, BD ⊥ AC, the perpendicular foot is the point D, CE ⊥ AB, and the perpendicular foot is the point E

The following results are proved: (1) BD ⊥ AC, CE ⊥ AB, ∠ AEC = ∠ ADB = 90 ° and ⊥ EAC = ∠ DAB, ⊥ AEC ⊥ ADB, ⊥ aead = acab, ⊥ aeac = ADAB, and ⊥ ead = ∠ cab, ⊥ ade ⊥ ABC; (2) in RT ⊥ AEC, ⊥ a = 60 °, ⊥ ace = 30 °, AC = 2ae, ⊥ ade ⊥

On the ellipse X & # 178 / 25 + Y & # 178 / 9 = 1, the distance from point P to the right guide line is equal to 4.5, and the distance from point P to the left focus is calculated

Eccentricity e = 4 / 5
The distance from point P to right focus is 4.5 × e = 18 / 5 [second definition of ellipse]
So the distance from P to the left focus is 10-18 / 5 = 32 / 5 [the first definition of ellipse]

RT angle ABC, angle B = 90 degrees, ab = 3cm, BC = 4cm, the triangle ABC fold, so that point C and point a coincide, get crease D

To find the perimeter of a triangle ADC, it is obvious that De is the perpendicular of AB, so ad = BD, AD + DC = BD + DC = 4cm, so the perimeter of a triangle ADC is 2 + 4 = 6cm,

Given the function f (x) = x ^ 2 + ainx, if G (x) = f (x) + 2 / X is a monotone function on [1, + ∞], find the value range of real number a

g'(x)=f'(x)+(2/x)=2x+a/x-2/x^2
Obviously when x is [1, + ∞]
At present, only a / X-2 / x ^ 2 ≥ 0 is required
The solution is a ≥ 2

If the length of the hypotenuse of a right triangle is 10 cm and the radius of the inscribed circle is 1 cm, then the circumference of the triangle is 10 cm
There are two options:
A：15cm B：22cm C：24cm D:26cm

The circumference of the triangle is 22cm
Option: B