Answer:
Step-by-step explanation:
B
2
3
Let the distance of two consecutive stones are x, x+1.
In ΔBCD, we have
tan60
o
=
x
h
⇒x=
3
h
.....(i)
In ΔABC, we have
tan30
o
=
x+1
h
⇒
3
1
=
x+1
h
⇒
3
h
+1=
3
h ......[from equation (i)]
⇒
3
2h
=1
⇒h=
2
3
km
solution
Lagrange multipliers:







(if

)

(if

)

(if

)
In the first octant, we assume

, so we can ignore the caveats above. Now,

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of

.
We also need to check the boundary of the region, i.e. the intersection of

with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force

, so the point we found is the only extremum.
Answer: It would be the first one, 0.001 kilometers per meter.
Step-by-step explanation: is 1,000 meters is equal to 1 kilometer, you would have to move the decimal on each until one is at one and the other is at 0.001. This would leave you with the first answer. Hope this helps!
Answer:
7
Step-by-step explanation:
Let the each leg be x so,
√(x²+x²)=√98
or, √(2x²)=√98
or, x√2=7√2
or, x=7
The slope of Line B is -1/5