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Alex
3 years ago
6

What is the slope of the line? A. -1 B. 0 C. 1 D. 20

Mathematics
2 answers:
LenaWriter [7]3 years ago
6 0
A<span>, Sketch of a heterogeneous plate, modelled as a cylindrically symmetric region that is thicker owing to a combination of greater depletion and lack of extension. </span>b<span>, Rock uplift at the edge and at the centre of the 'plateau' over 50 Myr of conductive relaxation for different values of </span>d<span> (</span>w = d<span>/10 and </span>h<span> = 100 km) and </span>Tb<span>, the initial basal temperature before mid-Tertiary time (modelled as </span><span>t = </span><span> 0), assumed to be the time of the thermal perturbation due to removal of the Farallon plate. The smaller symbols correspond to </span>Te<span> = 20 km and the larger to </span>Te<span> = 30 km (ref. </span>19). The observed range in residual rock uplift of the Colorado Plateau (Fig. 1d<span>) is indicated by the grey regions. </span>c<span>, Lateral migration of the 1,200 °C isotherm at </span><span>180-km depth as a function of </span>d<span> and </span>h<span> (results shown are for </span>Tb<span> = 1,300 °C). The slope of the lines represents the migration rate, which is a strong function of </span>w<span>, as illustrated by the lines with the smallest black symbols and fastest rate of encroachment: </span>d<span> = 800 km and </span>w = d<span>/100. All other results are shown for a larger value, </span>w = d<span>/10. Grey bars indicate rates observed in the migration of the onset of magmatism in </span>Fig. 2b<span>. </span>d<span>, Spatiotemporal pattern of rock uplift (every 5 Myr) for </span>d<span> = 800 km, </span>w = d<span>/100, </span>Te<span> = 30 km and </span>Tb<span> = 400 °C, showing higher values at the margins than in the interior and migration of the point of highest uplift slightly in from the margins of the plateau.</span>

sp2606 [1]3 years ago
3 0
The answer is B) 100%
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alexandr402 [8]

Answer:}

In polar coordinates P is

P  ( 12 ; 2.618 )  

Step-by-step explanation:

The point (ectangular coordinates)  

P ( -6√3  ;   6 )       P  (  x ; y )

Polar coordinates   P ( r  ;  θ )

x = r cos θ

y = r sin θ                  r > 0    and     0  ≤ θ ≤ 2π

Then

r  =  √ (x)² + (y)²        r  =  √(36)²*3 +( 36)²       r = √144

r  =  12   (hypothenuse module always positive)

The point   P  ( -6√3  ;   6 )   is in second cuadrant  between 90° and 180°

angle between  r  and horizontal axis x  is equal to θ

tan α =  l.opp/ l.adj.    tan θ =  y/x       tan θ = - 6 /6√3

tan α = - 1/√3

Then α  =  180⁰ - 150⁰  = 30⁰  or     θ =  150⁰

to express that value in radians we have :

1 π radian      =   180⁰         ⇒   3,1416   radians  =  180⁰

                                                      x ??                  =   150⁰

x =  radians

Finally the point is   P  ( 12 ; 2.618 )  

6 0
3 years ago
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