For the answer to the question above, the easiest way to determine is changing every runner's speed into the same unit.
<span>First = 10 m/s </span>
<span>Second = 10 miles/min = 16090.34 / 60 m/s (As 1 mile = 1609.34 meter and 1 min = 60 sec) </span>
<span>Second = 260.82 m/s </span>
<span>Third = 10 cm/hr = 10*(0.01)/60*60 (As 1 cm = 0.01 m and 1 hr = 60*60 sec) </span>
<span>Third = 0.000028 m/s </span>
<span>Fourth = 10 km/sec = 10*1000 m/s (As 1 km = 1000 m and time is already in sec) </span>
<span>Fourth = 10000 m/s </span>
<span>So fastest would be the one who covers the largest distance in 1 sec. It would be the fourth one.</span>
56, you are dividing by -2 each time
Answer:
y=-2
Step-by-step explanation:
Hello : let A(-3,-4) B(6,2)
the slope is : (YB - YA)/(XB -XA)
(2+4)/(6+3) = 6/9 = 2/3
an equation is the line is : y = ax+b a is a slope
y = (2/3)x+b
but this line passes by (6;2)
so : 2 = (2/3)(6)+b
b = -2
the equation is : y = (2/3)x-2
the the Y intercept of this line when : x= 0 so : y = (2/3)(0)-2
y=-2
Answer:
actually no triangle exists with the given angle measure. more than ONE unique triangle exists with the given angle measures. exactly one unique triangle exists with the given angle measures
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
