4 root 13 + 5 root 13-2 (root 13 + 1)

4 root 13 + 5 root 13-2 (root 13 + 1)

4√13+5√13-2(√13+1)
=9√13-2√13-2
=7√13-2

Let the inner angles a, B and C of △ ABC be opposite to a, B and C respectively. We know that a = 4, B = 5 and C = 61. (I) find the size of angle c; (II) find the area of △ ABC

(I) according to the meaning of the topic, a = 4, B = 5, C = 61, from the cosine theorem, we get COSC = 42 + 52 − (61) 22 × 4 × 5 = − 12, because C is the inner angle of the triangle, that is, ∠ C ∈ (0180 °), so ∠ C = 120 °; (6 points) (II) from (I) we get: sinc = sin120 ° = 32, and B = 5, a = 4, then the area of the triangle s △ ABC = 12b · a · sin120 ° = 12 × 5 × 4 × 32 = 53. (12 points)

It is known that the area of the triangle ABC is 10 times the root sign 3, a + B = 13, C is 60 degrees. Find the length of the three sides of the triangle

The perpendicular of BC through a intersects BC at point D
Because the angle c = 60 degrees
So high ad = B (radical 3) / 2
Area = a * B (radical 3) / 4 = 10 radical 3
ab=40
a+b=13
a=5,b=8
Ad = 4 radical 3
BD=5-4=1
C = radical (48 + 1) = 7
So the length of the three sides of the triangle is: a = 5; b = 8; C = 7

It is known that the straight line L passes through point a [1,1] and is only in the third quadrant. When the slope of L exists, what is the maximum value of the slope

Let the slope be K
y-1=k(x-1)
y=kx-k+1
When X0
When k > 0, K (x-1) + 1 decreases with the decrease of X, and must pass the third quadrant
So K cannot be greater than 0
When k = 0, the line y = 1 is not in the third quadrant
So, the maximum value of K is k = 0

x> 2, Y > 4, xy = 32, find the maximum value of log2 (x / 2) * log2 (Y / 4) and the X, y value at this time
True number in brackets, base number beside log

The original formula = (log2 (x) - 1) * (log2 (y) - 2), = (log2 (x) - 1) * (log2 (32 / x) - 2), = (log2 (x) - 1) * (5-log2 (x) - 2), = (log2 (x) - 1) * (3-log2 (x)), = - [log2 (x)] ^ 2 + 4log2 (x) - 3, (let t = log2 (x)) = - T ^ 2 + 4t-3, = - (T-2) ^ 2 + 1, because x > 2, t > 1 and T4)

The maximum value of y = log2 x + 4 / log2 x x belonging to [2,4] is

Let t = log2x
x∈[2,4]
Then t ∈ [1,2]
y(t)=t+4/t
y'(t)=1-4/t^2
Because t

Given that f (x) = log2 (x + 1), when the point (x, y) is on the f (x) image, the point (x / 3, Y / 2) is the point on the y = g (x) image, the analytic expression of G (x) is obtained

Let X / 3 = a, Y / 2 = B, then B = g (a)
Substituting x = 3A, y = 2B into f (x) = log2 (x + 1)
2b=log2(3a+1)
b=1/2 *log2(3a+1)
g(x)=1/2 *log2(3x+1)

If the image of the function y = log ∨ a (x + 3) - 1 (a > 0, a ≠ 1) passes through the point a, if the point a is on the straight line MX + NY + 1 = 0, m, n is greater than 0, find the minimum of 1 / n + 2 / m
minimum value

Log (a) x over fixed point (1,0),
Then y = log ∨ a (x + 3) - 1 passes the fixed point a (- 2, - 1)
Substituting MX + NY + 1 = 0, we get: - 2m-n + 1 = 0
2m+n=1.
∴1/n+2/m=(2m+n)/n+(4m+2n)/m
=(2m/n)+(2n/m)+5
≥ 4 + 5 = 9, (mean inequality)
The minimum value is 9

As shown in the figure, a particle moves in the first quadrant, x-axis and y-axis. One second later, it moves from the origin to (0,1), and then moves in the direction indicated by the arrow in the figure [that is, (0,0) → (0,1) → (1,1) → (1,0) →...] ], and move one unit length per second, then the coordinate of the particle position after 2011 second is (& nbsp;)
 

The particle moves once per second, and the number of seconds (0,0) → (0,1) → (1,1) → (1,0) is 1 second, 2 seconds, 3 seconds, 2 seconds to (1,1), 6 seconds to (2,2), 12 seconds to (3,3), 20 seconds to (4,4), and so on: to point (n, n), N2 + N seconds, ∵ when n = 44, n & # 178; + n = 44 &

A particle moves in the first quadrant, x-axis and y-axis. In the first second, it moves from the origin to (0,1), and then moves in the direction indicated by the arrow in the figure {that is, (0,0) - (0,1) - (1,1) - (1,0)} }And move one unit per second, then the coordinates of the particle position in the 35th second are ()
A. （4，0）B. （5，0）C. （0，5）D. （5，5）

It takes 3 seconds to reach (1,0), 4 seconds to reach (2,0), 4 unit lengths from (2,0) to (0,2), 4 + 4 = 8 seconds to reach (0,2), 9 seconds to reach (0,3), 6