Point J is on line segment IK. Given IK = 15 and JK = 6, determine the length IJ.
1 answer:
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11 = -3k - 22 - 8k
11 = -11k - 22 <em>added like terms (-3k and -8k)</em>
<u>+22</u> <u> +22 </u>
33 = -11k
-3 = k
Answer: k = -3
Let
x-------> t<span>he horizontal length of the roller coaster
y------> </span>the elevation of the roller coaster
c------> t<span>he incline of a roller coaster
</span>
we know that
applying the Pythagoras theorem
c²=x²+y²-----> equation 1
c=2y-----> equation 2
x=11+y----> equation 3
substitute equation 2 and equation 3 in equation 1
[2y]²=(11+y)²+y²
4y²=11²+22y+y²+y²
2y²-22y-121=0
using a graph tool-----> to resolve the second order equation
see the attached figure
the solution is
y=15.026 m
c=2*y-----> c=2*15.026----> c=30.052 m
the answer is
30.05 m
x-------> t<span>he horizontal length of the roller coaster
y------> </span>the elevation of the roller coaster
c------> t<span>he incline of a roller coaster
</span>
we know that
applying the Pythagoras theorem
c²=x²+y²-----> equation 1
c=2y-----> equation 2
x=11+y----> equation 3
substitute equation 2 and equation 3 in equation 1
[2y]²=(11+y)²+y²
4y²=11²+22y+y²+y²
2y²-22y-121=0
using a graph tool-----> to resolve the second order equation
see the attached figure
the solution is
y=15.026 m
c=2*y-----> c=2*15.026----> c=30.052 m
the answer is
30.05 m