The length of the transverse axis of a hyperbola is given by the expression ( 2a ), where ( a ) is the distance from the center of the hyperbola to each vertex along the transverse axis. For a hyperbola centered at the origin with the standard form ( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 ) (horizontal transverse axis) or ( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 ) (vertical transverse axis), the value of ( a ) determines the extent of the transverse axis. Thus, the transverse axis length varies directly with ( a ).
The length of the transverse axis of a hyperbola is determined by the value of (2a), where (a) is the distance from the center to each vertex along the transverse axis. In the standard forms of hyperbolas, such as ((x-h)^2/a^2 - (y-k)^2/b^2 = 1) or ((y-k)^2/a^2 - (x-h)^2/b^2 = 1), (a) represents this distance. Therefore, to find the length of the transverse axis, you would use the expression (2a).
2*pi*sqrt(L/g) this expression gives (approximately) the period (in seconds) of a pendulum whose length is L (metres) and g is the acceleration due to gravity = 9.8 metres/second2.
73
You cannot solve this single equation. You can either change the subject so that it gives x = 12/y or xy = 12, which is the equation of a rectangular hyperbola.
length of arc/length of circumference = angle at centre/360 Rearranging the equation gives: length of arc = (angle at centre*length of circumference)/360
a - b
The length of the transverse axis of a hyperbola is determined by the value of (2a), where (a) is the distance from the center to each vertex along the transverse axis. In the standard forms of hyperbolas, such as ((x-h)^2/a^2 - (y-k)^2/b^2 = 1) or ((y-k)^2/a^2 - (x-h)^2/b^2 = 1), (a) represents this distance. Therefore, to find the length of the transverse axis, you would use the expression (2a).
A triangle has 3 line segments
tan 13/30
2*pi*sqrt(L/g) this expression gives (approximately) the period (in seconds) of a pendulum whose length is L (metres) and g is the acceleration due to gravity = 9.8 metres/second2.
the area of a rectangle with width x and length 6x is 6x^2 what does the coefficient 6 mean in terms of the problem
It gives the velocity of the object in the radial direction. The graph gives no information whtsoever about motion in a transverse direction.
The question gives an expression, not an equation. An expression cannot have a solution.
Not necessarily. "Amplitude" gives you an idea about how "strong" a wave is; the concept applies both to longitudinal and to transverse waves.
73
What is its area? Divide the perimeter by 4: that gives you the length of one side. Then square that length: that gives you the area.
You cannot solve this single equation. You can either change the subject so that it gives x = 12/y or xy = 12, which is the equation of a rectangular hyperbola.