Solve for y in the first equation.
Let the radius of the circle be r. Then the line from the external point through the center of the circle which extends to the far point on the circle has length 3r .By the tangent - secant theorem
t^2 = 3r * r = 3r^2 ( where t is the length of the tangent).
So t = √(3r^2) = √3r answer.
V^2/(1-v^2/c^2)=R
v^2=R(1-v^2/c^2)
v^2=R-Rv^2/c^2
v^2-Rv^2/c^2=R
v^2(1-R/c^2)=R
v=sqrt(R/(1-R/c^2))
where R was original right side, dont forget plus minus
Answer:
See the proof below
Step-by-step explanation:
Let the line AB be a straight line on the parallelogram.
A dissection of the line (using the perpendicular line X) gives:
AY ≅ BX
Another way will be using the angles.
The angles are equal - vertically opposite angles
Hence the line AY ≅ BX (Proved)
Answer:
x=(k√x)/(w^2)(cube√z)
x varies directly with square root of y is:
x = k √y
And inversely with the square of w and cube root of s is:
/(w^2)(cube√z)
Now put those together:
x=(k√x)/(w^2)(cube√z)
The formula is:
x=ky/wz but add the specific info like square, roots, or etc
