What's the value of 2 root sign 3-2

What's the value of 2 root sign 3-2

=2×1.732-2
=3.464-2
=1.464

What is 2 / 2 + 2 times the root 3 What is 2 + 2 times the root 2 / 3?

2/(2+2√3) =1/(1+√3) =(√3 -1)/[(1+√3)(√3 -1)]=(√3 -1)/[√3² -1²]=(√3 -1)/2

If sin α = the root of 5, what is the value of Cos4 α

sina=√5/5
So (Sina) ^ 2 = (√ 5 / 5) = 1 / 5
So cosa = ± √ (1 - (Sina) ^ 2) = ± 2 √ 5 / 5
cos4α=1-2(sin2α)^2
sin2α=2sinacosa=±4/5
therefore
cos4α=1-2(sin2α)^2=1-2X16/25=-7/25

If sin α = the root of 5, then the value of Cos4 α is?

cos2a=1-2(sina)^2=3/5
cos4a=2(cos2a)^2-1=2*9/25-1=-7/25

Sin α is known= Five 5, then the value of SiN4 α - Cos4 α is 0______ .

sin4α-cos4α
=sin2α-cos2α
=2sin2α-1
=-3
5，
So the answer is: - 3
5．

Given the points P (3,1), m (5,1), n (0, - 1-radical 3), the straight line L passes through the point P and intersects with the line segment Mn, so as to find the value range of L inclination angle The answer is [0,30 degrees] u [45 degrees, 180 degrees]

Let the function formula of the line l be: y = KX + B when the straight line passes through points P and N, the following two equations can be obtained: 1 = k * 3 + B ① - 1 - √ 3 = k * 0 + B ② by substituting formula ①, we get: 1 = k * 3-1 - √ 3K = (2 + √ 3) / 3, Tan α = k = (2 + √ 3) / 3 α = 51.09 ° when the straight line passes through points P and m, we can get the following two equations: 1 = k * 5 + B ③ the solution is obtained by the

It is known that two fixed points a (2,5), B (- 2,1), m and N are two moving points on the line L passing through the origin, | Mn | = 2 times the root sign 2 It is known that two fixed points a (2,5), B (- 2,1), m and N are two moving points on the line L passing through the origin, | Mn | = 2 times the root sign 2, l is parallel to ab. if the intersection point C of the straight line am and BN is on the Y axis, find the coordinates of points m, N and C

Let k = 1 (slope) straight line L cross the origin, let C (0, y) m (a, a) n (C, c) connect BC AC, ab = 4, root sign 2 Mn / AB = CN / Ca = cm / CB = 1 / 2 CN ^ 2 = C ^ 2 + (c-y) ^ 2 Ca ^ 2 = 4 + (5-y) ^ 2cm ^ 2 = a ^ 2 + (a-y) ^ 2 CB ^ 2 = 4 + (1-y) ^ 2Mn ^ 2 = (C-A) ^ 2 + (C-A) ^ 2 = 8 C-A = 2 to get a, C, y

It is known that two fixed points a (2,5), B (- 2,1), m (in the first quadrant) and N are two moving points on the line L passing through the origin, and Mn = 2, root sign 2, l ∥ ab. if the intersection point C of the line am and BN is on the Y axis, find the coordinates of point C

Firstly, the equation of L is written according to the known conditions
x/(-2-2)=y/(1-5)
So the equation of L is:
Y=x
Let m coordinate (x, x) because m is y = x on L
The coordinates of n points can be obtained as (X-2, X-2) or n '(x + 2, x + 2) by trigonometric geometry
Am equation:
(X-2)/(2-x)=(Y-5)/(5-x) …………………… （1）
BN equation:
(X+2)/(x-2+2)=(Y-1)/(x-2-1) …………………… （2）
Or BN 'equation
(X+2)/(x+2+2)=(Y-1)/(x+2-1) ………………………… （3）
Let x = 0 solve x = 1, y = - 3
Combine (1) (3) and let x = 0 solve x = - 1 (do not conform to m point in the first quadrant, round off)
So m coordinates are (1,1), n coordinates are (- 1, - 1) C coordinates are (0, - 3)

Given that the straight line y = X-1 intersects with hyperbola at two points m, the abscissa of the midpoint of line Mn in n is - 2 / 3, and the focus of hyperbola C is root 7. The hyperbolic equation is solved There are points to pursue

Let hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1
m,n (x1,y1) (x2,y2)
Then Y1 = x1-1
y2=x2-1
x1^2/a^2-y1^2/b^2=1
x2^2/a^2-y2^2/b^2=1
(x1+x2)(x1-x2)/a^2-(y1+y2)(y1-y2)/b^2=0
Because X1 + x2 = - 4 / 3
y1+y2=x1+x2-2=-10/3
y1-y2=x1-x2
So 4 / A ^ 2 = 10 / b ^ 2
Point C is the root 7
a^2+b^2=7
therefore
x^2/2-y^2/5=1

Given m = 1 / 3, n = 1 / 27, find the value of (m-n / radical m-radical n) + (M + 4n-4 radical Mn / radical m-2n) Exercises for the first semester of Grade 8 (Shanghai) Try the answer to the fourth question in the seventh industry

Root number (n-4m) + n-4m
=(root number m + root number n) + (root number m-2 root number n)
=2 radical m-radical n
=2 root 3 / 3 - radical 3 / 9
=Radical 3 / 3