The golden ratio, phi occurs many places in the platonic solids. The dihedral angle on the dodecahedron is 2*atan(phi), and the dihedral angle on the icosahedron is 2*atan(phi2) or 2*atan(phi + 1). The mid radius of the dodecahedron is similarly phi2/2 or (phi + 1)/2, and the mid radius on the icosahedron is phi/2. There are several other measures within Platonic solids which involve phi.
An angle does not have a mid point.
Normally it's its centre which is at the mid point of its diameter.
mid condition! Its a confusing play on words!!
It is the mid-point of a line segment that has end points
well,first the radius is half of the chord. Radius is the distance from the circle centre to the chord end. The chord is the line joining the ends of the arc. Draw this line. Call the distance from the arc of the circle at its deepest point to the mid point of the chord "c". If extended, this line will go throught the centre of the circle. Call half the length of the chord "y". Then the properties of circles and chords is that c(d-c)=y2 where d is the circle diameter, so that d = y2/c + c. And then radius is half that.
In a regular polygon, the apothem is a line from the centre to the mid-point of one of the flat sides. The radius is a line from the centre to a corner, which is longer.
False. The Earth's radius and surface area do not increase as new oceanic crust is formed at mid-oceanic ridges. Instead, the creation of new crust at mid-oceanic ridges is balanced by the destruction of older crust at subduction zones, maintaining the overall size of the Earth.
A narrow chord blade weighs less because of the less massive cross section but their vibration frequency is too low to operate successfully without the use of a mid span shroud to provide added stiffness. Elimination of the mid span shroud increases the fan performance/efficiency. RR was one of the first to design a successful wide chord blade and got around the weight penalty by developing a partially hollow titanium blade design. P&W followed suit. GE chose to solve the weight penalty by developing epoxy/graphite fan blades.
To Determine the depth of the sulcus , six sites are probed. There are three sites on the facial, including mesiofacial, mid facial, and disto facial, and three sites on the lingual, including mesiolingual, mid lingual, and disto lingual.
Using a compass determine the direction of south. Using a clock find out the time when the sun is south. For every hour after mid-day you are 15 degrees west of the point where the sun is south at mid-day. The clock should be set to mid-day when the sun is at longitude zero to get absolute values rather than relative values.
No, Earth's radius and surface area are not increasing to accommodate new oceanic crust. The process of seafloor spreading at mid-ocean ridges is balanced by subduction, where older crust is recycled back into the mantle. This maintains the overall size of Earth and its surface area.
thye pulled rocks from a drill and studied them obviously
The golden ratio, phi occurs many places in the platonic solids. The dihedral angle on the dodecahedron is 2*atan(phi), and the dihedral angle on the icosahedron is 2*atan(phi2) or 2*atan(phi + 1). The mid radius of the dodecahedron is similarly phi2/2 or (phi + 1)/2, and the mid radius on the icosahedron is phi/2. There are several other measures within Platonic solids which involve phi.
False. The Earth's radius and surface area are not increasing to accommodate new oceanic crust. Instead, the process of seafloor spreading at mid-ocean ridges involves the creation of new oceanic crust, which is balanced by the destruction of crust at subduction zones, maintaining the Earth's overall surface area.
The mid point of the diameter is the circle's centre: Mid point of (2, -3) and (8, 7) is ((2 + 8)/2, (-3 + 7)/2) = (5, 2) The radius is half the diameter, but as the equation for a circle requires the radius squared: radius = diameter/2 → radius² = diameter² / 4 Diameter (and diameter²) can be found using Pythagoras: diameter² = change_in_x² + change_in_y² = (8 - 2)² + (7 - -3)² = 6² + 10² = 136 → radius² = 136/4 = 34 Circle with centre (X, Y) and radius r has equation: (x - X)² + (y - Y)² = r² → (x - 5)² + (y - 2)² = 34 → x² -10x + 25 + y² - 4y + 4 = 34 → x² + y² - 10x - 4x - 5 = 0
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