How to calculate (1 + 0.25 + 0.39) x (0.25 + 0.39 + 0.87) - (1 + 0.25 + 0.39 + 0.87) x (0.25 + 0.39),

How to calculate (1 + 0.25 + 0.39) x (0.25 + 0.39 + 0.87) - (1 + 0.25 + 0.39 + 0.87) x (0.25 + 0.39),

(1+0.25+0.39)X(0.25+0.39+0.87)-(1+0.25+0.39+0.87)X(0.25+0.39)=(1+0.25+0.39)X(0.25+0.39+0.87)-(1+0.25+0.39）X(0.25+0.39)+0.87X(0.25+0.39)=(1+0.25+0.39)X(0.25+0.39）+(1+0.25+0.39)X0.87-(1+0.25+0.39）X(0...
How to calculate 0.73-0.25-0.73 + 0.25 in a simple way
0.73-0.25-0.73+0.25
=(0.73-0.73)+(0.25-0.25)
=0+0
=0
A simple method to calculate 5.2 times 10.1 2.33 times 0.5 times 0.4 times 5
5.2x10.1=5.2x(10+0.1)=52+0.52=52.52
2.33x0.5x0.4x5=2.33x0.5x2=2.33x1=2.33
The second formula multiplies the last two terms first!
What's 1.2 percent
1.2/100=12/1000=3/250
Given the vector a = (cos20 ° sin20 ° and B = (0, - 1), what is the angle between a and B
a=（cos20°,sin20°）,b=(0,-1),
|a|=1,|b|=1
a.b=-sin20°
cos=(-sin20°)/1*1=-sin20°=cos110°
So the angle between vectors a and B is 110 degrees
∵cos=a*b/|a||b|=-sin20º=cos110º
∴=110º
|a|=1,|b|=1
ab=cos20°X0+sin20°X(-1)=-sin20°
Let the angle between vector a and B be x, then there is:
cosx=ab/|a||b|=(-sin20°)/1
cosx=-sin20°=-cos70°=cos110°
So: x = 110 degree
tanB=(sin20+1)/cos20
sinB/cosB=(sin20+1)/cos20
sinBcos20=cosBsin20+cosB
sinBcos20-cosBsin20=cosB
sin(B-20)=cosB
B-20=90-B
2B=110
B=55
Do not know whether to calculate the direction, ha ha, this is a bit forgotten
The direction is - 55 degrees
[(SiNx + xcosx) (1 + cosx) + X (SiNx) ^ 2] / (1 + cosx) ^ 2 the detailed process of getting the result, thank you!
[(sinX+XcosX)(1+cosX)+X(sinX)^2]/(1+cosX)^2
The result of this question should be: (SiNx + x) / (1 + cosx)
[(sinX+XcosX)(1+cosX)+X(sinX)^2]/(1+cosX)^2
=[(sinX+XcosX)(1+cosX)+X(1-cosX^2)]/(1+cosX)^2
=[(sinX+XcosX)(1+cosX)+X(1+cosX)(1-cosX)]/(1+cosX)^2
=[(sinX+XcosX)+X(1-cosX)]/(1+cosX)
=(sinX+XcosX+X-XcosX)/(1+cosX)
=(sinX+X)/(1+cosX)
Note: SiNx ^ 2 + cosx ^ 2 = 1
[(sinX+XcosX)(1+cosX)+X(sinX)^2]/(1+cosX)^2
=[(sinx+xcosx)(1+cosx)+x(1-cos^2x)]/(1+cosX)^2
=[(sinx+xcosx)+x(1-cosx)]/1+cosx
=(sinx+x)/(1+cosx)
85 percent is equal to several hundred percent, 75 percent is equal to several percent
17/20
3/4
If two points a (cos40 ° sin40 ° and B (sin20 ° cos20 ° are known), then the quadratic value of AB vector is
AB = OB-OA= (sin20°-cos40°,cos20°-sin40°)|AB|^2 = AB.AB= (sin20°-cos40°,cos20°-sin40°).(sin20°-cos40°,cos20°-sin40°)=(sin20°-cos40°)^2+(cos20°-sin40°)^2= 2- 2(sin20°cos40°+cos20°sin4...
Let f (x) = (AX + b) SiNx + (Cx + D) cosx, try to determine the constants a, B, C, D, such that f '(x) = xcosx
We have known that f ′ (x) = [(AX + b) SiNx + (Cx + C + X + D) the known f ′ (x) = [(AX + b) SiNx] '= [(AX + b) SiNx]' = [(AX + b) SiNx] '= [(AX + b) SiNx]' + [(Cx + b) cosx] '= [(AX + b) SiNx]' + [(Cx + b) cosx] '= [(AX + b) SiNx + [(Cx + b) cosx]' - [(AX + C (x + b) cosx (as the known known f ′ (x + b) ′ (x + b) the known f ′ (x) \\\\\(x (x) = x cosx, and then another another another one of the other of the other of the final final final final final final final final final of the final of the new cosx of the new new cosx d = 0 − C = 0A A = D = 1, B = C = 0
What's the ratio of 3.8 to 12
3.8/12=19/60
3.8/12 = 38 / 120, about 2 = 19 / 60
3.8 = 38 / 100 = 19 / 5 19 / 5 vs. 12 = 19 / 60