The value of the algebraic expression 2cos10 °− sin20 ° cos20 ° is () A. 2B. 3C. 1D. 12

The value of the algebraic expression 2cos10 °− sin20 ° cos20 ° is () A. 2B. 3C. 1D. 12

2cos10 °− sin20 ° cos20 ° = 2cos (300 − 20 °) − sin20 ° cos20 ° = 3cos200cos200 = 3, so select B
Find the value of (2cos10 ° minus sin20 °) divided by cos20 °
(2cos10°-sin20°)/cos20°=[2cos(30°-20°)-sin20°]/cos20° =[2cos30°cos20°+2sin30°sin20°)-sin20°]/cos20° =[(√3cos20°+sin20°)-sin20°]/cos20° =√3cos20°/cos20° =√3
3 / (sin20 'squared) - 1 / (cos20' squared) + 64 (sin20 'squared) = as follows
3/sin20°-1/cos20°+ 64sin20° =4(3cos20-sin20)/(2sin20cos20) +64sin20 =4(1+2cos40)/sin40 +64sin20 =8(cos60+cos40)/sin40 + 64sin20 =16(cos50cos10)/sin40 + 64sin20 =16sin80/sin40 +64sin20 =32cos40 +32(1-cos40) =32
How to use calculator, I want to calculate logarithm and exponent and so on
Log followed by the base, and then add the real number. Index with the base, and then add a few power
Given a function y = (2m-1) x - (n + 3), when m and N are the values, this function is also a positive proportion function, when m and N are the values, y = (2m-1) x - (n + 3)
The value of x increases with the increase of X, and intersects with y axis in the negative half axis
If M = 1, n = 2, find the coordinates of the intersection of the function image and x-axis and y-axis
1. When 2m-1 ≠ 0, N + 3 = 0, i.e. m ≠ 1 / 2, n = - 3, the linear function is a positive proportional function. When 2m-1 > 0, N + 3 > 0, i.e. m > 1 / 2, n > - 3, the value of Y increases with the increase of x, and intersects with y axis at negative half axis 3, M = 1, n = 2, y = X-5, x = 0, y = - 5; when y = 0, x = 5. Therefore, the intersection coordinate of function image and X axis is (5,0
As shown in the figure, it is known that the image of the first-order function y = KX + B (k is not equal to 0) intersects with the x-axis and y-axis at two points a (1,0) and B (0, - 1), respectively, and intersects with the image of the inverse scale function y = m / X (M is not equal to 0) at point C in the first quadrant
(1) 2m-1≠0 m≠1/2
n+3=0 n=-3
When m ≠ 1 / 2 and N = - 3, the linear function is also a positive proportional function
(2) 2m-1>0 m>1/2
n+3>0 m>-3
When m > 1 / 2, n > - 3, the value of Y increases with the increase of X, and intersects with y axis in the negative half axis
(3) m=1,n=2
y=x-5
When y = 0, X-5 = 0, x = 50 m > 1 / 2
n+3>0 m>-3
When m > 1 / 2, n > - 3, the value of Y increases with the increase of X, and intersects with y axis in the negative half axis
(3) m=1,n=2
y=x-5
When y = 0, X-5 = 0, x = 5
Intersection with X axis (5,0)
When x = 0, y = 0-5 = - 5
The point of intersection (0, - 5) with y axis is retracted
2log5 (10) + log5 (0.25) =? Urgent
LOG5(100)+LOG5(0.25)=LOG5(25)=2
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Given the function y = (M + 1) x + (M2-1), when m takes what value, y is a linear function of X, and when m takes what value, y is a positive proportional function of M
Y is a linear function of X
The coefficient of X is not equal to 0
m+1≠0
m≠-1
Y is a positive proportional function of M
The constant term is zero
The coefficient of X is not equal to 0
m+1≠0
m2-1=0
So m = 1
Given that log5 is the logarithm of base 3 = a, log5 is the logarithm of base 2 = B, find log25 as the logarithm of base 12
Log25 is the bottom 12
=log5(12)/log5(25)
=log5(12) /2
=log5(2*2*3)/2
=[log5(2)+log5(2)+log5(3)]/2
=(2a+b)/2
log25（12）=log5*5（12）=0.5log5（12）=0.5（log5（3）+log5（4））=0.5（a+2b）
0.5a+b
（2a+b）/2
log25(12)=log25(3)+log25(4)=0.5log5(3)+log5(2)=0.5a+b
Given Z-M + y, M is a constant, y is a positive proportional function of X, when x = 2, z = 1, when x = 3, z = - 1, find the functional relationship between Z and X
Y is a positive proportional function of X, so let y = KX
Z=M+KX
Substitute x = 2, z = 1; X = 3, z = - 1
2K+M=1,3K+M=-1
K=-2,M=5
The functional relation between Z and X: z = - 2x + 5
What is the logarithm power of 1 / 2 with 2 as the base and 3 as the base
A kind of Let y = (1 / 2) ^ [log]_ 2 (3)]log_ (1/2) (y) = log_ 2 (3) log (y) / log (1 / 2) = log (3) / log (2), the bottom formula log (y) / - log (2) = log (3) / log (2) - log (y) = log (3) log (1 / Y) = log (3) 1 / y = 3Y = 1 / 3 ■ (1 / 2