Question 1075691
Let {{{ t = 1 }}} = time in hrs the planes have been flying
Let {{{ r[1] }}} = the rate in mi/hr of the slower plane
{{{ r[1] + 140 }}} = the rate in mi/hr of the faster plane
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{{{ ( r[1]*t )^2 + ( ( r[1] + 140 )*t )^2 = 260^2 }}}
{{{ ( r[1]*1 )^2 + ( ( r[1] + 140 )*1 )^2 = 260^2 }}}
{{{ r[1]^2 + ( r[1] + 140 )^2 = 260^2 }}}
{{{ r[1]^2 + r[1]^2 + 280r[1] + 19600 = 67600 }}}
{{{ 2r[1]^2 + 280r[1] - 48000 = 0 }}}
{{{ r[1]^2 + 140r[1] - 24000 = 0 }}}
{{{ r[1] = ( -b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 1 }}}
{{{ b = 140 }}}
{{{ c =-24000 }}}
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{{{ r[1] = ( -140 +- sqrt( 140^2-4*1*(-24000) ))/(2*1) }}}
{{{ r[1] = ( -140 +- sqrt( 19600 + 96000 )) / 2 }}}
{{{ r[1] = ( -140 +- sqrt( 115600 )) / 2 }}}
{{{ r[1] = ( -140 + 340 ) / 2 }}}
{{{ r[1] = 200/2 }}}
{{{ r[1] = 100 }}}
and
{{{ r[1] + 140 = 100 + 140 }}}
{{{ r[1] + 140 =240 }}}
Their rates are:
100 mi/hr and
240 mi/hr
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check:
{{{ ( r[1]*t )^2 + ( ( r[1] + 140 )*t )^2 = 260^2 }}}
{{{ ( 100*1 )^2 + ( 240*1 )^2 = 260^2 }}}
{{{ 10000 + 57600 = 67600 }}}
{{{ 67600 = 67600 }}}
OK