Answer:
tan(2u)=[4sqrt(21)]/[17]
Step-by-step explanation:
Let u=arcsin(0.4)
tan(2u)=sin(2u)/cos(2u)
tan(2u)=[2sin(u)cos(u)]/[cos^2(u)-sin^2(u)]
If u=arcsin(0.4), then sin(u)=0.4
By the Pythagorean Identity, cos^2(u)+sin^2(u)=1, we have cos^2(u)=1-sin^2(u)=1-(0.4)^2=1-0.16=0.84.
This also implies cos(u)=sqrt(0.84) since cosine is positive.
Plug in values:
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.84-0.16]
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.68]
tan(2u)=[(0.4)(sqrt(0.84)]/[0.34]
tan(2u)=[(40)(sqrt(0.84)]/[34]
tan(2u)=[(20)(sqrt(0.84)]/[17]
Note:
0.84=0.04(21)
So the principal square root of 0.04 is 0.2
Sqrt(0.84)=0.2sqrt(21).
tan(2u)=[(20)(0.2)(sqrt(21)]/[17]
tan(2u)=[(20)(2)sqrt(21)]/[170]
tan(2u)=[(2)(2)sqrt(21)]/[17]
tan(2u)=[4sqrt(21)]/[17]
Answer:what grade is this for?
Step-by-step explanation:
?
It is not a circle but you can find the equation of the ellipse.
To do this we need to work out the major and minor radii and the centre
The centre is at (9, 7)
The major (y) radius is 1 and the minor (x) radius is 5
Therefore the equation is (x-9)/5 + (y-7)/1 = 1
Assuming the h is not in the exponent, R would equal The square root of V over 3.14h
X - y = 9
x + 2y = 27
You can subtract the top equation from the bottom
3y = 18
y = 6
x = 15
The two numbers are 15 and 6
