<span>N(t) = 16t ; Distance north of spot at time t for the liner.
W(t) = 14(t-1); Distance west of spot at time t for the tanker.
d(t) = sqrt(N(t)^2 + W(t)^2) ; Distance between both ships at time t.
Let's create a function to express the distance north of the spot that the luxury liner is at time t. We will use the value t as representing "the number of hours since 2 p.m." Since the liner was there at exactly 2 p.m. and is traveling 16 kph, the function is
N(t) = 16t
Now let's create the same function for how far west the tanker is from the spot. Since the tanker was there at 3 p.m. (t = 1 by the definition above), the function is slightly more complicated, and is
W(t) = 14(t-1)
The distance between the 2 ships is easy. Just use the pythagorean theorem. So
d(t) = sqrt(N(t)^2 + W(t)^2)
If you want the function for d() to be expanded, just substitute the other functions, so
d(t) = sqrt((16t)^2 + (14(t-1))^2)
d(t) = sqrt(256t^2 + (14t-14)^2)
d(t) = sqrt(256t^2 + (196t^2 - 392t + 196) )
d(t) = sqrt(452t^2 - 392t + 196)</span>
-49/15 -3 4/15...........
Answer:
√(23)
Step-by-step explanation:
The Pythagorean Theorem is a² + b² = c²
so...
a = 7
b = 3
7² + 3² = c²
14 + 9 = c²
√(23) = c
PLEASE RATE!! I hope this helps!!
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Answer:
x=13
Step-by-step explanation:
First, you add the x's together: 2x+x=3x
Now you have:3x-1=38
Add 1 to both sides: 3x=39
1 cancel on the left side of the equation
Divide 3 on both sides: x=39/3
3 cancels on the left side and 39 divided by 3 is 13:
x=13
Hope this helps!
Hypotenuse = 5√6
Let the length of each leg be x, since it is Isosceles .
x² + x² = (5√6)²
2x² = (5 * 5 * √6 * √6)
2x² = 150
x² = 150/2
x² = 75
x = √75
x = √(25 * 3)
x = √25 * √3
x = 5√3
